Chemistry & Vibration Modes

Working out why chemists like character tables, from first principles

When I first met finite groups and their representations in an undergraduate algebra class, they felt magical but esoteric. Differential equations felt connected to the real world; character tables did not. So it came as a genuine surprise when my chemist friends started casually using them — not as curiosities, but as working tools, on the same page as obviously practical concerns, like bond angles and vibrational spectra.

I never sat down to figure out what they were actually doing. I knew, vaguely, that the symmetry of a molecule controlled its behavior, and then that the character table was how this control was encoded. But how did symmetry affect behavior? This post is my attempt to work it out from first principles.

The goal is concrete: starting from Lagrangian mechanics of a molecule near an equilibrium, we will derive the procedure chemists use to read vibrational mode structure off a character table. Along the way the various ingredients — configuration space, the kinetic metric, the molecular point group, the isotypic decomposition — will show up in the order the physics needs them.

Because I’m a mathematician, I’m going to keep things needlessly general. Instead of narrowing our focus on the real world, we work on a general Riemannian manifold $(X,g)$ with isometry group $G = \operatorname{Isom}(X,g)$. This way we can clearly see where the geometry, the analysis and the group theory interact.

The setup

We will work with the simplest classical toy model of a molecule: atoms are point masses in our ambient space $(X, g)$, attracted to and repelled by one another through forces derivable from a potential energy. There is no electronic structure, no quantum mechanics — just classical particle dynamics on a Riemannian manifold. This is a less-than-honest model of real molecules, but it is enough to recover the chemists’ character-table procedure, which is what I am trying to understand.

A molecule, then, is a finite collection of point masses in $(X,g)$, each atom labeled by a mass $m_i > 0$, with no two atoms ever occupying the same point. Ordered atom positions are tracked as a point in $X^n$, but atoms of the same mass are physically indistinguishable, so swapping two of them gives the same molecule. The honest configuration space is the quotient

$$\widehat Q \;:=\; \bigl(X^n \setminus \Delta\bigr) \big/ \Sigma,$$

where $\Delta$ is the diagonal (configurations with two atoms coincident) and $\Sigma \subset S_n$ is the group permuting atoms within each mass class. Motions of the molecule in space are curves in $\widehat Q$.

Two structures on $\widehat Q$ drive everything that follows.

First, the kinetic metric. A moving molecule has kinetic energy $\tfrac12 \sum_i m_i\, g(\dot x_i, \dot x_i)$, which is a quadratic form on $T\widehat Q$. To unpack: a tangent vector at the configuration $(x_1, \ldots, x_n)$ is a tuple $(v_1, \ldots, v_n)$ with $v_i \in T_{x_i} X$ — one velocity vector per atom — and

$$\widehat{\mathbf g}\bigl((v_1, \ldots, v_n),\, (w_1, \ldots, w_n)\bigr) \;=\; \sum_{i=1}^n m_i\, g(v_i, w_i),$$

the mass-weighted sum of the per-atom inner products. Written as a symmetric 2-tensor on $Q$, we’d say

$$\widehat{\mathbf g}=\bigoplus_i m_i g$$

Its $\Sigma$- and $G$-invariance are immediate: permuting equal-mass atoms leaves it alone, and moving the whole molecule by an isometry of $X$ leaves it alone.

Second, the potential $V \colon \widehat Q \to \mathbb R$. We require $V$ to be $G$-invariant, where $G$ acts on $\widehat Q$ by simultaneously moving every atom by the same isometry of $X$ (this is the diagonal action of $G$ on $X^n$, which descends to $\widehat Q$ since it commutes with $\Sigma$). Geometrically, the energy of a molecule does not depend on where in $(X,g)$ it sits or how it is oriented. In practice $V$ depends only on pairwise geodesic distances and bond angles, which are $G$-invariant by construction.

The Lagrangian is $\mathcal{L} = \tfrac12\, \widehat{\mathbf g}(\dot q, \dot q) - V(q)$, and the Euler–Lagrange equations take the coordinate-free form

$$\nabla_{\dot q} \dot q \;=\; -\operatorname{grad}_{\widehat{\mathbf g}} V,$$

where $\nabla$ is the Levi–Civita connection of $\widehat{\mathbf g}$.

Everything in the rest of this post is a consequence of these few ingredients: the manifold $\widehat Q$, the kinetic metric $\widehat{\mathbf g}$, the $G$-invariant potential $V$, and the induced $G$-action on $\widehat Q$.

Linearizing at an equilibrium

An equilibrium is a point $\widehat e \in \widehat Q$ with $dV_{\widehat e} = 0$ — a configuration at which no net force acts on any atom. We want to study small motions near such an equilibrium, and the natural first step is to linearize the Euler–Lagrange equation at $\widehat e$.

Consider a one-parameter family of trajectories near the equalibrium $e$. We might parameterize such a family in tersm of a variable $s$, say $q(s,t)$ with $q(0,t) \equiv \widehat e$ the equalibirum itself. Let $\xi(t) = \partial_s q|_{s=0}$ be the small displacement at first order in $s$. Since $s=0$ is the equalibirum, for all $t$ this vector based at $e$, so $\xi(t)$ is a curve in $T_{\widehat e}\widehat Q$. This curve captures the infinitesimal behavior of our family of solutions near $e$: since a vector at $e$ is really a collection of velocity vectors on each atom of our molecule, we can think of the path $\xi(t)$ as an animation of changing velocity vectors on our equalibrium configuration, or an infinitesimal motion.

Our goal is to find the linearized equation that $\xi(t)$ obeys. We get it by differentiating the Euler–Lagrange equation $\nabla_{\dot q}\dot q \;=\; -\operatorname{grad}_{\widehat{\mathbf g}}V$. Two vector fields on the image of $q(s, t)$ organize the calculation. The first is $T \;=\; \partial_t q,$ which points along the solutions in our family — at each point it is the velocity of the trajectory passing through, obtained by moving in $t$ at fixed $s$. The second is $S \;=\; \partial_s q$, which points across the family, from one solution to its neighbor — it is the displacement you see by moving in $s$ at fixed $t$.

Thus, the fact that our family $q(s,t)$ is a family of solutions, means that for each $s$ the Euler lagrangue equation is satisfied by

$\nabla_T T = -\operatorname{grad}_{\widehat {\mathbf g}} V$

From this we want to extract an equation for $\xi = S|_{s=0}$. The idea is to covariantly differentiate both sides of the Euler–Lagrange equation along the variation direction $S$, then evaluate at $s = 0$. Written out, we want to compute

$$\underbrace{\nabla_S\, (\nabla_T T)\big|_{s=0}}_{\text{left side}} \;=\; \underbrace{-\,\nabla_S\, (\operatorname{grad} V)\big|_{s=0}}_{\text{right side}}$$

and see what it says about $\xi$. Each side turns out to be something we already care about. Two facts about this setup will do all of the work.

• Since $S$ and $T$ are derivatives of coordinate functions, $[S, T] = 0$.
• At $s = 0$ the trajectory $q(0, t) \equiv \widehat e$ is constant in $t$, so $T(0, t) = 0$ for all $t$.

Left side

Rearranging the definition of the Riemann curvature tensor of $\widehat{\mathbf g}$,

$$\nabla_S \nabla_T T \;=\; \nabla_T \nabla_S T \,+\, \nabla_{[S, T]}\, T \,+\, R(S, T)\, T.$$

The bracket term $\nabla_{[S, T]}\, T$ vanishes, since $[S, T] = 0$. So

$$\nabla_S \nabla_T T \;=\; \nabla_T \nabla_S T \,+\, R(S, T)\, T.$$

Next, we can swap the $\nabla_S T$ for $\nabla_T S$. Precisely, the torsion free identity for the Levi Civita connection reads

$$\nabla_S T \,-\, \nabla_T S \,-\, [S, T] \;=\; 0.$$

The bracket term again vanishes, leaving $\nabla_S T = \nabla_T S$. Substituting,

$$\nabla_S \nabla_T T \;=\; \nabla_T \nabla_T S \,+\, R(S, T)\, T.$$

Now evaluate at $s = 0$. The curvature term vanishes because one of its arguments is $T(0, t) = 0$ (and its applied to $T(0,t)=0$!). What remains is $\nabla_T \nabla_T S|_{s=0}$. Along the constant curve $t \mapsto \widehat e$, covariant differentiation reduces to ordinary differentiation in the vector space $T_{\widehat e}\widehat Q$, and $S(0, t) = \xi(t)$. So putting it all together

$$\underbrace{\nabla_S\, (\nabla_T T)\big|_{s=0}}_{\text{left side}} \;=\;\nabla_T \nabla_T S \big|_{s=0} \;=\; \ddot \xi(t) \;\in\; T_{\widehat e}\widehat Q.$$

Right side

At $s = 0$, $S(0, t) = \xi(t)$ and $q(0, t) = \widehat e$, so $\nabla_S$ applied to any vector field along the family reduces at $s = 0$ to $\nabla_\xi$ at $\widehat e$. In particular,

$$\nabla_S\, (\operatorname{grad} V)\big|_{s=0} \;=\; \nabla_\xi\, \operatorname{grad} V\,\big|_{\widehat e}.$$

The map $\xi \mapsto \nabla_\xi \operatorname{grad} V|_{\widehat e}$ is linear in $\xi$ (because the covariant derivative is linear in its lower index), so it is a linear operator on $T_{\widehat e}\widehat Q$. We give it a name:

$$\mathcal H \colon T_{\widehat e}\widehat Q \to T_{\widehat e}\widehat Q, \qquad \mathcal H\, \xi \;:=\; \nabla_\xi \operatorname{grad} V\,\big|_{\widehat e}.$$

The right side of the Euler–Lagrange equation, linearized at the equilibrium, is $-\mathcal H\, \xi$.

Computing $\mathcal H$

We have named the operator, but at this point all we know about it is that it is linear. To get a formula — and to uncover whatever further structure $\mathcal H$ has — we probe it by pairing $\mathcal H \xi$ against an arbitrary tangent vector $\eta$ via the kinetic metric. From the metric compatibility of the Levi–Civita connection,

$$\begin{aligned} \widehat{\mathbf g}(\mathcal H\, \xi, \eta) &\;=\; \widehat{\mathbf g}(\nabla_\xi \operatorname{grad} V, \eta) \\ &\;=\; \xi\bigl(\widehat{\mathbf g}(\operatorname{grad} V, \eta)\bigr) \,-\, \widehat{\mathbf g}(\operatorname{grad} V,\, \nabla_\xi \eta). \end{aligned}$$

Evaluating at $\widehat e$ does two things to this expression. The second term vanishes because $\operatorname{grad} V|_{\widehat e} = 0$. The first term simplifies via the defining property of the gradient, $\widehat{\mathbf g}(\operatorname{grad} V, \eta) = dV(\eta) = \eta(V)$. Together,

$$\widehat{\mathbf g}_{\widehat e}(\mathcal H\, \xi, \eta) \;=\; \xi\bigl(\eta(V)\bigr)\,\big|_{\widehat e}.$$

The right-hand side $\xi(\eta(V))|_{\widehat e}$ has a hidden symmetry that we can extract by playing $\xi$ and $\eta$ off against each other. Using the bracket identity for vector fields acting on functions,

$$\xi(\eta(V)) \,-\, \eta(\xi(V)) \;=\; [\xi, \eta](V) \;=\; dV([\xi, \eta]),$$

and the fact that $dV_{\widehat e} = 0$, we conclude $\xi(\eta(V))|_{\widehat e} = \eta(\xi(V))|_{\widehat e}$. But $\eta(\xi(V))|_{\widehat e}$ is exactly what our formula returns with the roles of $\xi$ and $\eta$ reversed: $\widehat{\mathbf g}_{\widehat e}(\mathcal H\, \eta,\, \xi) \;=\; \eta(\xi(V))\,\big|_{\widehat e}$. Chaining these together,

$$\begin{aligned} \widehat{\mathbf g}_{\widehat e}(\mathcal H\, \xi,\, \eta) &\;=\; \xi(\eta(V))\,\big|_{\widehat e} \\ &\;=\; \eta(\xi(V))\,\big|_{\widehat e} \\ &\;=\; \widehat{\mathbf g}_{\widehat e}(\mathcal H\, \eta,\, \xi) \\ &\;=\; \widehat{\mathbf g}_{\widehat e}(\xi,\, \mathcal H\, \eta), \end{aligned}$$

where the last step uses symmetry of the metric. Comparing the first and last entries is exactly the statement that $\mathcal H$ is self-adjoint with respect to the kinetic metric, a crucial property of $\mathcal H$ for our future computational work.

To turn the implicit equation $\widehat{\mathbf g}_{\widehat e}(\mathcal H\, \xi,\, \eta) = \xi(\eta(V))|_{\widehat e}$ into an explicit formula for $\mathcal H \xi$ as a vector, we apply the musical isomorphism $\sharp\colon T^*\widehat Q \to T\widehat Q$ — the map that takes any 1-form $\omega$ to the unique vector $\omega^\sharp$ with $\widehat{\mathbf g}(\omega^\sharp, Y) = \omega(Y)$. The 1-form here is $\eta \mapsto \xi(\eta(V))|_{\widehat e}$, and

$$\mathcal H\, \xi \;=\; \bigl(\eta \mapsto \xi(\eta(V))\big|_{\widehat e}\bigr)^{\sharp}.$$

This coordinate-free expression is somewhat abstract; the same defining relation also lets us compute $\mathcal H$ in any basis we like. Pick a basis $\{e_a\}$ of $T_{\widehat e}\widehat Q$ and plug $\xi = e_a$, $\eta = e_b$ into the formula. We get a number — call it $H_{ab}$ — given by

$$H_{ab} \;:=\; \widehat{\mathbf g}_{\widehat e}\bigl(\mathcal H\, e_a,\, e_b\bigr) \;=\; e_a\bigl(e_b(V)\bigr)\,\big|_{\widehat e}.$$

If we use a coordinate basis, $e_a = \partial/\partial x^a$ and $e_b = \partial/\partial x^b$, the right-hand side is a familiar second partial derivative

$$H_{ab} \;=\; \frac{\partial^2 V}{\partial x^a\, \partial x^b}\bigg|_{\widehat e}.$$

So $H$ is just the matrix of second partial derivatives of $V$ at the equilibrium.

But $H$ is not the matrix of our operator $\mathcal H$ itself: $\mathcal H e_a$ sits inside a metric pairing on the left of the defining formula, so the kinetic metric is tangled with our operator. To peel off the metric and extract the matrix of $\mathcal H$ alone, we use self-adjointness.

Treat $\widehat g$, $H$, and $\mathcal H$ as ordinary matrices in our basis. For column vectors $\xi$ and $\eta$,

$$\widehat{\mathbf g}_{\widehat e}(\mathcal H\, \xi,\, \eta) \;=\; (\mathcal H \xi)^T\, \widehat g\, \eta \;=\; \xi^T\, \mathcal H^T \widehat g\, \eta,$$

and self-adjointness of $\mathcal H$ tells us $\mathcal H^T \widehat g = \widehat g\, \mathcal H$. Setting the result equal to $H(\xi, \eta) = \xi^T H \eta$ gives us the matrix identity

$$\xi^T H \eta = \xi^T \widehat g \mathcal H \eta,$$

and demanding it hold for all $\xi$ and $\eta$ implies

$$H \;=\; \widehat g\, \mathcal H , \qquad \text{equivalently} \qquad \mathcal H \;=\; g^{-1}\, H.$$

The matrix of the operator $\mathcal H$ in any basis is the inverse kinetic-metric matrix times the matrix of second partial derivatives of $V$.

Putting it all together

We set out to study an infinitesimal deformation of the equilibrium solution $q(t) \equiv \widehat e$ — a curve $\xi(t) \in T_{\widehat e}\widehat Q$ describing the small motion away from rest. By covariantly differentiating both sides of the Euler–Lagrange equation along the variation direction $S$ and evaluating at $s = 0$, the relation

$$\nabla_S \nabla_T T \;=\; -\,\nabla_S \operatorname{grad} V$$

became a linear second-order ODE on the tangent space at the equilibrium,

$$\ddot \xi \;+\; \mathcal H\, \xi \;=\; 0,$$

where $\mathcal H$ is the self-adjoint operator built directly from the two structures we started with — the (invese of the) kinetic metric and the (second derivatives of the) potential. Thus, the entire linear theory of molecular vibration is captured by the qualtiative behavior of ODE’s of this type.

A simplified derivation in $\mathbb R^n$

For comparison, here is the linearization in flat space, where the calculation is much shorter. Take $X = \mathbb R^d$, identify $\widehat Q$ with an open subset of $\mathbb R^N$ (where $N = n d$) once we pick a basis, treat the kinetic metric as a constant symmetric positive-definite matrix $M$, and let $V \colon \mathbb R^N \to \mathbb R$ be the potential. The Lagrangian is

$$\mathcal L \;=\; \tfrac{1}{2}\, \dot q^{\,T} M\, \dot q \;-\; V(q),$$

and the Euler–Lagrange equation reads

$$M\, \ddot q \;=\; -\nabla V(q).$$

Linearize around an equilibrium $\widehat e$ (so $\nabla V(\widehat e) = 0$). Write $q(t) = \widehat e + \xi(t)$ and Taylor-expand the gradient,

$$\nabla V(\widehat e + \xi) \;=\; H\, \xi \;+\; O(\xi^2),$$

where $H = \nabla^2 V(\widehat e)$ is the matrix of second partial derivatives at the equilibrium. Plug in, drop the $O(\xi^2)$ terms, and multiply by $M^{-1}$:

$$\ddot \xi \;+\; \mathcal H\, \xi \;=\; 0, \qquad \mathcal H \;=\; M^{-1}\, H.$$

Solving the linear theory

We are now ready to use the structure of $\mathcal H$ to solve $\ddot \xi + \mathcal H\, \xi = 0$.

Modes by sign of eigenvalue

Because $\mathcal H$ is self-adjoint with respect to the kinetic metric, the spectral theorem gives us a $\widehat{\mathbf g}_{\widehat e}$-orthonormal basis of $T_{\widehat e}\widehat Q$ consisting of eigenvectors of $\mathcal H$, with real eigenvalues.

Pick any eigenvector $v$ with eigenvalue $\omega^2$, and look for solutions of the form $\xi(t) = c(t)\, v$ where $c$ is a real-valued function of time. Plugging in,

$$\ddot c\, v \,+\, \mathcal H(c\, v) \;=\; (\ddot c \,+\, \omega^2 c)\, v,$$

so $\xi(t) = c(t)\, v$ is a solution exactly when $c(t)$ satisfies the scalar ODE

$$\ddot c \,+\, \omega^2 c \;=\; 0.$$

The behavior depends entirely on the sign of $\omega^2$.

Positive eigenvalue ($\omega^2 > 0$). The equation is a harmonic oscillator,

$$c(t) \;=\; A \cos(\omega t) \,+\, B \sin(\omega t).$$

The mode wobbles back and forth in the $v$ direction with frequency $\omega$ — bounded oscillation. These are the normal modes of the molecule, and the corresponding $\omega$ are the normal frequencies.

Negative eigenvalue ($\omega^2 < 0$). Writing $\omega^2 = -\lambda^2$ with $\lambda > 0$, the equation becomes $\ddot c = \lambda^2 c$, with general solution

$$c(t) \;=\; A\, e^{\lambda t} \,+\, B\, e^{-\lambda t}.$$

For generic initial conditions the $e^{\lambda t}$ piece dominates: an arbitrarily small displacement in the $v$ direction grows exponentially in time. The equilibrium is unstable along $v$ — geometrically, $\widehat e$ is a saddle of $V$ in this direction, the potential going down rather than up.

Zero eigenvalue ($\omega^2 = 0$). The equation reduces to $\ddot c = 0$, with general solution

$$c(t) \;=\; A \,+\, B\, t.$$

The molecule drifts at constant velocity in the $v$ direction. There is no restoring force, no oscillation, no exponential growth — just uniform motion. These zero modes are neither oscillations nor instabilities; they are something else, and a structural part of the story.

Putting the modes together. Decomposing an arbitrary initial condition into the eigenbasis, the general solution to $\ddot \xi + \mathcal H \xi = 0$ is

$$\xi(t) \;=\; \sum_\alpha c_\alpha(t)\, v_\alpha,$$

where each $c_\alpha$ evolves independently according to its eigenvalue $\omega_\alpha^2$. The equilibrium $\widehat e$ is stable — every small perturbation stays small — exactly when every $\omega_\alpha^2 > 0$, i.e., when $\mathcal H$ is positive definite. Equivalently, the matrix of second partials of $V$ at $\widehat e$ is positive definite, and $\widehat e$ is a local minimum of $V$ rather than a saddle or a maximum.

Real molecules sit at local minima of their potential, so we expect every nonzero $\omega_\alpha^2$ to be positive — every nonzero mode an oscillation. But zero eigenvalues are not avoidable: there are always zero modes, forced by a structural feature of $\widehat Q$ that we have so far ignored.

Where the zero modes come from

The structural feature we have so far ignored is that the equilibrium $\widehat e$ is never isolated. Equilibria of $V$ on $\widehat Q$ always come in families, because $V$ is $G$-invariant.

If $V(g \cdot q) = V(q)$ for every $g \in G$, then carrying a critical point along the $G$-action gives another critical point, and the full orbit

$$\widehat{\mathcal O} \;:=\; G \cdot \widehat e \;\subset\; \widehat Q$$

consists entirely of equilibria. This is not a pathology; it is the mathematical shadow of a fact we already accept — rigidly translating or rotating an equilibrium molecule gives another equilibrium molecule. The ambient isometry group acts trivially on the energy, so it acts non-trivially on the space of equilibria.

The tangent directions to this orbit at $\widehat e$ form a linear subspace

$$N \;:=\; T_{\widehat e} \widehat{\mathcal O} \;\subset\; T_{\widehat e}\widehat Q.$$

These are the infinitesimal rigid motions: velocities generated by a one-parameter subgroup of $G$ acting on the whole molecule at once.

Along any such direction the potential is constant — it has to be, because the orbit consists entirely of equilibria and they all share the same energy. So $\mathcal H$ kills $N$:

$$\mathcal H\, \xi \;=\; 0 \qquad \text{for all } \xi \in N.$$

These are exactly the zero-eigenvalue modes from the previous section. We saw what they do under the linearized dynamics: they don’t oscillate, they drift. The molecule “oscillates” along these directions at zero frequency — which is to say, it just coasts off, rigidly translating or rotating through space.

These zero modes are not interesting as vibrations. They are the imprint on $T_{\widehat e}\widehat Q$ of the ambient isometry group, pure and simple. The genuine vibrational content of the molecule — the oscillatory modes we set out to compute — must live in the complementary directions.

The vibrational subspace

We need to separate the rigid-motion directions from the honest vibrations. The natural way is to take the orthogonal complement of $N$ in $T_{\widehat e}\widehat Q$ with respect to our kinetic metric $\widehat{\mathbf g}_{\widehat e}$ — the only inner product on $T_{\widehat e}\widehat Q$ available to us. Define

$$\mathcal V \;:=\; N^{\perp_{\widehat{\mathbf g}}} \;\subset\; T_{\widehat e}\widehat Q,$$

so that the tangent space splits orthogonally as

$$T_{\widehat e}\widehat Q \;=\; N \,\oplus\, \mathcal V.$$

The space $\mathcal V$ is where the real vibrational dynamics lives. Before going further it is worth checking that the operator $\mathcal H$ behaves nicely on the splitting $T_{\widehat e}\widehat Q = N \oplus \mathcal V$ — that it sends each summand into itself, and that the restriction is still self-adjoint.

That $\mathcal H$ preserves $\mathcal V$ is a consequence of self-adjointness. For any $\xi \in \mathcal V$ and any $\eta \in N$,

$$\widehat{\mathbf g}_{\widehat e}(\mathcal H\, \xi,\, \eta) \;=\; \widehat{\mathbf g}_{\widehat e}(\xi,\, \mathcal H\, \eta) \;=\; 0,$$

since $\mathcal H$ kills $N$. So $\mathcal H \xi$ pairs trivially with every element of $N$, which is to say $\mathcal H \xi \in N^\perp = \mathcal V$. Self-adjointness of the restriction $\mathcal H|_{\mathcal V}$ then comes for free: the restriction of a self-adjoint operator to an invariant subspace, together with the restricted inner product, is again self-adjoint. So $\mathcal H|_{\mathcal V}$ is a self-adjoint operator on the inner product space $(\mathcal V, \widehat{\mathbf g}_{\widehat e}|_{\mathcal V})$.

Restricted to $\mathcal V$ the operator is also non-degenerate (assuming $\widehat e$ is sufficiently generic — that we haven’t accidentally produced extra zero directions transverse to the orbit). The linearized Euler–Lagrange equation then decouples cleanly:

$$\ddot \xi_0 \;=\; 0, \qquad \ddot \xi_\perp \;+\; \mathcal H|_{\mathcal V}\, \xi_\perp \;=\; 0,$$

where $\xi = \xi_0 + \xi_\perp$ with $\xi_0 \in N$ and $\xi_\perp \in \mathcal V$. Zero modes drift; vibrational modes oscillate; the two never talk to each other.

So the problem reduces to a concrete finite-dimensional eigenvalue problem: diagonalize the self-adjoint operator $\mathcal H|_{\mathcal V}$ on a vector space of dimension $n\dim X - \dim N$. For a molecule of $n$ atoms in three-space, $\mathcal V$ is generically of dimension $3n - 6$ — which for even modestly sized molecules is already a substantial matrix.

Self-adjointness gives us a lot for free, even before we touch the matrix entries. The spectral theorem on $(\mathcal V, \widehat{\mathbf g}_{\widehat e}|_{\mathcal V})$ yields an orthogonal direct sum

$$\mathcal V \;=\; \bigoplus_\lambda E_\lambda$$

over the (real) eigenvalues $\lambda$ of $\mathcal H|_{\mathcal V}$, where $E_\lambda$ is the corresponding eigenspace. Each $\lambda$ is a squared vibrational frequency $\omega^2$, and the modes oscillating at frequency $\omega$ span $E_\lambda$. This is the most physical decomposition of $\mathcal V$ available: split the modes by frequency.

Computing the eigenvalues and eigenvectors of $\mathcal H|_{\mathcal V}$ explicitly requires the specific potential $V$ and a real diagonalization. But there are coarser questions about the spectrum that we might hope to answer without solving the eigenvalue problem at all:

• How many distinct vibrational frequencies are there?
• How many independent vibrational modes oscillate at each frequency?

Equivalently: how many distinct eigenvalues $\lambda$ are there, and what are the dimensions $\dim E_\lambda$? This is coarse data about the eigendecomposition — the shape of the orthogonal direct sum, not the actual eigenvalues. And it is exactly the kind of information that symmetry alone can pin down, without reference to the specific $V$.

This is where the group theory comes to save the day.

Symmetry and representation theory

Why symmetry?

We already know quite a bit about $\mathcal H|_{\mathcal V}$. It is a self-adjoint operator on a finite-dimensional inner product space, so the spectral theorem hands us an orthonormal basis of eigenvectors with real eigenvalues, and the small motions of the molecule decompose into independently oscillating modes accordingly. But this is a generic structure, available for any self-adjoint operator on any finite-dimensional inner product space. It tells us the modes exist; it tells us nothing about the organization of the spectrum — how many distinct frequencies, with what multiplicities, are forced.

If we want to say more about the spectrum without knowing the specifics of the potential $V$, we need a structural input beyond ”$\mathcal H$ is self-adjoint.” Symmetry is a natural candidate: a group acting on $\mathcal V$ that commutes with $\mathcal H|_{\mathcal V}$ would constrain its spectrum, in cases where the symmetry is rich enough. So our plan in this section is to identify a relevant group of symmetries of our linearized system, prove that it commutes with $\mathcal H|_{\mathcal V}$, and read off the consequences.

The symmetries of an equilibrium

The natural source of symmetries is the ambient isometry group $G = \mathrm{Isom}(X, g)$, which we have already met as the group preserving everything we have built. We are not interested in all of $G$, though, only the part that fixes our particular equilibrium $\widehat e$. Set

$$P \;:=\; G_{\widehat e} \;=\; \bigl\{\, h \in G \;:\; h \cdot \widehat e = \widehat e \,\bigr\}.$$

What does an element of $P$ look like concretely? Pick a representative $(x_1, \ldots, x_n) \in X^n$ for the equivalence class $\widehat e \in \widehat Q$. Then $h \cdot \widehat e = \widehat e$ in $\widehat Q$ exactly when the diagonally-moved tuple $(h \cdot x_1, \ldots, h \cdot x_n)$ is $\Sigma$-equivalent to $(x_1, \ldots, x_n)$. In other words: $h$ is a rigid motion of $X$ whose effect on the molecule’s atoms is the same as a permutation of equal-mass atom labels — there exists $\sigma_h \in \Sigma$ with $h \cdot x_i = x_{\sigma_h(i)}$ for every $i$.

So $P$ is the group of rigid motions of $X$ that send the molecular shape to itself. Each $h \in P$ comes with a permutation $\sigma_h \in \Sigma$, and $h \mapsto \sigma_h$ is a group homomorphism $P \to \Sigma$.

In examples in $X = \mathbb{R}^3$:

• Water. $P$ has 4 elements: the molecular plane (containing all three atoms) is a mirror; the line through the oxygen perpendicular to the H–H line is a $\pi$-rotation axis; and one further mirror containing that rotation axis is perpendicular to the molecular plane. The rotation and one mirror swap the two H’s; the molecular-plane mirror and the identity do not. Abstractly, $P$ is the Klein four-group $\mathbb{Z}/2 \times \mathbb{Z}/2$.
• Ammonia. $P$ has 6 elements: two non-trivial rotations by $\pm 2\pi/3$ about the nitrogen axis (cycling the three H’s), and three mirror planes containing that axis. Abstractly, $P$ is the dihedral group $D_3 \cong S_3$ — the symmetry group of an equilateral triangle, or equivalently permutations of the three H atoms.
• Methane. $P$ has 24 elements — every symmetry of a regular tetrahedron, realized in $\mathbb{R}^3$ as rotations and reflections that permute the four equivalent H atoms.

Action on the tangent space, and a classification

Each $h \in G$ acts on the configuration space $\widehat Q$ as a diffeomorphism

$$\Phi_h \colon \widehat Q \to \widehat Q, \qquad q \;\mapsto\; h \cdot q,$$

obtained by applying the isometry $h$ to every atom of the configuration. For $h \in P$, $\Phi_h$ fixes the basepoint $\widehat e$, so its differential at $\widehat e$ is a well-defined linear map of the tangent space. Collecting these differentials into a single homomorphism gives a representation of $P$:

$$\rho \colon P \;\to\; \mathrm{GL}\bigl(T_{\widehat e}\widehat Q\bigr), \qquad h \;\mapsto\; d(\Phi_h)_{\widehat e}.$$

The representation has special structure built into how it was constructed: $P$ acts on $\widehat Q$ by isometries of $\widehat{\mathbf g}$ (since $P \subset G = \mathrm{Isom}(X,g)$, and the descended action is isometric), and the differential of an isometry at a fixed point is itself a linear isometry of the tangent inner product space. So $\rho$ takes values in the orthogonal group of the kinetic-metric inner product:

$$\rho \colon P \;\to\; \mathrm{O}\bigl(T_{\widehat e}\widehat Q,\, \widehat{\mathbf g}_{\widehat e}\bigr).$$

We also note that $\rho$ preserves the splitting $T_{\widehat e}\widehat Q = N \oplus \mathcal V$. The orbit $\widehat{\mathcal O} = G \cdot \widehat e$ is $G$-invariant as a set, hence $P$-invariant, so $\rho(h)$ preserves $N = T_{\widehat e}\widehat{\mathcal O}$; preserving the metric, it then preserves $\mathcal V = N^\perp$ as well. So $\rho$ restricts to an isometric representation $\rho|_{\mathcal V} \colon P \to \mathrm{O}(\mathcal V, \widehat{\mathbf g}_{\widehat e}|_{\mathcal V})$ on the vibrational subspace.

What kind of group is $P$? A priori $\rho(P)$ sits inside the (large) orthogonal group $\mathrm O(T_{\widehat e}\widehat Q) \cong \mathrm O(n \dim X)$. But the abstract group $P$ — sitting inside the much smaller Lie group $G = \mathrm{Isom}(X, g)$ — has a more constrained shape than that.

Warmup in $\mathbb R^d$. When $X = \mathbb R^d$ this is easy to see directly. Place the origin at the (mass-weighted) center of mass of the molecule. Every $h \in P$ permutes equal-mass atoms among themselves, so it preserves the center of mass and fixes the origin. The isometries of $\mathbb R^d$ fixing the origin are exactly $\mathrm{O}(d)$, so

$$P \;\hookrightarrow\; \mathrm{O}(d).$$

For $X = \mathbb R^3$ this is $\mathrm{O}(3)$. The symmetry group of any molecular equilibrium in $\mathbb R^3$ is a subgroup of $\mathrm{O}(3)$.

This is a sharp constraint. Sometimes $P$ is infinite — a linear molecule like $\mathrm{CO}_2$ or $\mathrm{HCN}$ has $P$ containing a continuous $\mathrm{SO}(2)$ of rotations about the molecular axis. But if the molecule is not collinear, $P$ is finite. The finite subgroups of $\mathrm{O}(3)$ are completely classified. The finite subgroups of $\mathrm{SO}(3)$ are the cyclic groups $\mathbb{Z}/n$, the dihedral groups $D_n$ (of order $2n$), and the three Platonic rotation groups $A_4$ (tetrahedral), $S_4$ (octahedral, equivalently the rotations of a cube), and $A_5$ (icosahedral, equivalently the rotations of a dodecahedron); the finite subgroups of $\mathrm{O}(3)$ are obtained from these by adjoining orientation-reversing elements like reflections.

So in $\mathbb R^3$ we recover exactly the symmetry classification chemists already use: the molecular point groups. The same list in chemistry notation reads $C_n$, $D_n$, $T$, $O$, $I$ for the rotation-only groups, with their reflection-extended cousins $C_{nv}$, $C_{nh}$, $S_n$, $D_{nh}$, $D_{nd}$, $T_d$, $T_h$, $O_h$, $I_h$.

The general case. The center-of-mass argument is genuinely Euclidean: in a curved Riemannian manifold the global Riemannian center of mass need not exist, and there is no canonical “origin” of $X$ to place. So we need a different route in the general setting — and fortunately the same kind of conclusion can be reached more abstractly, by identifying $P$ as a compact subgroup of $G$ and then applying a structural theorem about Lie groups.

That $P$ is compact follows from a standard fact in Riemannian geometry: the action of $G = \mathrm{Isom}(X, g)$ on $X$ is proper, meaning that the (setwise) stabilizer of any compact subset of $X$ is itself compact in $G$. The atom positions $\{x_1, \ldots, x_n\}$ form a compact (finite) subset of $X$, and any $h \in P$ permutes these atoms among themselves — so $P$ is contained in the setwise stabilizer of $\{x_1, \ldots, x_n\}$, which is compact by properness. Hence $P$ is compact.

(In typical cases $P$ is in fact finite. The precise condition is that the atoms are not all contained in a codimension-2 totally geodesic submanifold of $X$ — informally, that the molecule is “spread out enough” in $X$ that no continuous family of rotations preserves it. In $\mathbb R^3$ this is the non-collinear case, since codimension 2 means a single geodesic line; a collinear molecule like $\mathrm{CO}_2$ has all its atoms on one line, and $P$ then contains a continuous $\mathrm{SO}(2)$ of rotations about that line, giving $\mathrm{O}(2) = C_{\infty v}$ or $D_{\infty h}$ — still compact, just no longer finite. In higher-dimensional or differently shaped ambient spaces the condition adapts: in $\mathbb R^4$, a coplanar configuration similarly admits a continuous rotational stabilizer, and so on.)

Now apply a theorem of E. Cartan (the Cartan–Iwasawa–Mal’cev theorem): every compact subgroup of a Lie group $G$ is contained in some maximal compact subgroup of $G$, and any two maximal compacts are conjugate. So $P$ embeds in a maximal compact of $G$.

For Riemannian symmetric spaces this maximal compact has a clean identification, depending on type. For non-compact-type spaces (like $\mathbb H^d$) and Euclidean-type spaces ($\mathbb R^d$ itself), the maximal compact of $G$ is exactly the point stabilizer at any chosen basepoint — so $P$ ends up inside that point stabilizer, which is $\mathrm{O}(d)$ in either case. For compact-type spaces (like $S^d$), $G$ is itself compact and serves as its own maximal compact; $P$ sits inside all of $G$, which is a larger rotation group ($\mathrm O(d+1)$ for $S^d$). Either way $P$ is a compact subgroup of a finite-dimensional rotation group.

In Thurston-geometry terms: $\mathbb R^3$ and $\mathbb H^3$ give $\mathrm O(3)$; $S^3$ gives $\mathrm O(4)$; the smaller geometries ($\mathrm{Nil}$, $\mathrm{Sol}$, $\mathbb H^2 \times \mathbb R$, $\widetilde{\mathrm{SL}_2}$) give correspondingly smaller maximal compacts inside $\mathrm O(3)$. In every case $P$ is a compact subgroup of a familiar rotation group, and for sufficiently spread-out molecules — those whose atoms are not all on a codimension-2 totally geodesic submanifold of $X$ — it is in fact finite, falling into the classification we already wrote down.

For the rest of the post we assume our equilibrium satisfies this generic condition (in $\mathbb R^3$, that the molecule is non-collinear), so $P$ is a finite subgroup of $\mathrm{O}(d)$ — and the representation theory of $P$ is the classical theory of finite groups.

$\mathcal H$ is $P$-equivariant

Intuition. Think of $\mathcal H$ as encoding the molecule’s linear restoring force: a small displacement $\xi$ from equilibrium produces a restoring force proportional to $-\mathcal H \xi$. Now suppose we displace not by $\xi$ but by its symmetry-image $\rho(h)\xi$, for some $h \in P$. By symmetry of the molecule, the restoring force on this rotated/reflected displacement should be the same rotation/reflection of the original restoring force — that is, $\rho(h)$ applied to $-\mathcal H \xi$. So

$$\mathcal H \bigl(\rho(h)\xi\bigr) \;\stackrel{?}{=}\; \rho(h)\bigl(\mathcal H \xi\bigr)$$

— the operator $\mathcal H$ commutes with $\rho(h)$. Behind the scenes, this works because every ingredient of $\mathcal H$ — the potential $V$, the kinetic metric $\widehat{\mathbf g}$, and the equilibrium $\widehat e$ — is $P$-invariant.

Proof. Recall how $\mathcal H$ was defined: it is the unique linear operator on $T_{\widehat e}\widehat Q$ satisfying

$$\widehat{\mathbf g}_{\widehat e}(\mathcal H\xi,\, \eta) \;=\; \xi\bigl(\eta(V)\bigr)\big|_{\widehat e} \qquad \text{for all } \xi, \eta \in T_{\widehat e}\widehat Q$$

— in other words, $\mathcal H$ is built from the bilinear form $\xi(\eta(V))|_{\widehat e}$ via the kinetic metric. To show $\mathcal H$ commutes with $\rho(h)$, it suffices to show that both sides of this defining relation transform compatibly under $\rho$. The metric is $\rho$-invariant by construction — $\rho(h)$ is a linear isometry. The new fact we need is that the bilinear form $\xi(\eta(V))|_{\widehat e}$ is $\rho$-invariant too. This follows from a single chain-rule fact:

Chain rule at a critical point. Let $f \colon M \to \mathbb R$ be smooth and $\Phi \colon M \to M$ a diffeomorphism with $\Phi(p) = p$. If $df_p = 0$, then for any tangent vectors $\xi, \eta$ at $p$,

$$\xi\bigl(\eta(f \circ \Phi)\bigr)\big|_p \;=\; \bigl(d\Phi_p\, \xi\bigr)\Bigl(\bigl(d\Phi_p\, \eta\bigr)(f)\Bigr)\Big|_p.$$

The slogan: at a critical point of $f$, the second-derivative bilinear form of $f \circ \Phi$ is $f$‘s second-derivative form with both inputs replaced by their pushforwards through $d\Phi_p$.

We will apply the chain rule with $f = V$, $\Phi = \Phi_h$, $p = \widehat e$, and $d\Phi_p = \rho(h)$. Combined with the metric invariance and the defining relation for $\mathcal H$, this gives the equivariance directly. Apply each fact in turn to the expression $\widehat{\mathbf g}_{\widehat e}(\mathcal H \rho(h)\xi,\, \rho(h)\eta)$:

$$\begin{aligned} \widehat{\mathbf g}_{\widehat e}\bigl(\mathcal H \rho(h)\xi,\, \rho(h)\eta\bigr) &\;=\; \bigl(\rho(h)\xi\bigr)\Bigl(\bigl(\rho(h)\eta\bigr)(V)\Bigr)\Big|_{\widehat e} && \text{(defining relation, with } \rho(h)\xi, \rho(h)\eta) \\ &\;=\; \xi\bigl(\eta(V \circ \Phi_h)\bigr)\big|_{\widehat e} && \text{(chain rule, applied to } V \circ \Phi_h\text{)} \\ &\;=\; \xi\bigl(\eta(V)\bigr)\big|_{\widehat e} && \text{(}V \circ \Phi_h = V\text{)} \\ &\;=\; \widehat{\mathbf g}_{\widehat e}\bigl(\mathcal H\xi,\, \eta\bigr) && \text{(defining relation, with } \xi, \eta) \\ &\;=\; \widehat{\mathbf g}_{\widehat e}\bigl(\rho(h)\mathcal H\xi,\, \rho(h)\eta\bigr) && \text{(metric invariance).} \end{aligned}$$

The first and last expressions are pairings against the same vector $\rho(h)\eta$, and they agree for every $\eta$. By non-degeneracy of the kinetic metric,

$$\mathcal H \rho(h)\xi \;=\; \rho(h)\mathcal H\xi.$$

Conclusion.

The operator $\mathcal H$ is $P$-equivariant: $\mathcal H \circ \rho(h) = \rho(h) \circ \mathcal H$ for every $h \in P$.

This is the structural fact we will use everywhere from here on.

Rep theory to the rescue

Recall where we are. The spectral theorem gave us the orthogonal eigendecomposition $\mathcal V = \bigoplus_\lambda E_\lambda$ from the self-adjointness of $\mathcal H|_{\mathcal V}$, and we posed two coarse questions about it: how many distinct eigenvalues are there, and what is the dimension of each eigenspace $E_\lambda$? The new structural input we have is that $\mathcal{V}$ carries a representation of $P$, and the operator $\mathcal H|_{\mathcal V}$ “plays well” with the $P$-representation $\rho|_{\mathcal V}$ (they commute).

The first consequence of this commutation is that each eigenspace of $\mathcal H$ is itself a representation of $P$. Take an eigenvector $v$ with eigenvalue $\lambda$. For any $h \in P$,

$$\mathcal H\bigl(\rho(h)\, v\bigr) \;=\; \rho(h)\bigl(\mathcal H\, v\bigr) \;=\; \rho(h)\bigl(\lambda v\bigr) \;=\; \lambda\, \rho(h)\, v,$$

so $\rho(h)\, v$ is also an eigenvector of $\mathcal H$ with the same eigenvalue $\lambda$. The eigenspace $E_\lambda$ is therefore preserved by every $\rho(h)$, and so $E_\lambda$ inherits the action and becomes a representation of $P$ in its own right.

To go further, we use a key fact from rep theory.

Building blocks: irreducible representations

The first thing rep theory tells us is that any representation of a finite group splits into “indivisible” pieces. An irreducible representation (or irrep) is one with no proper $P$-invariant subspace — the smallest possible representation.

Maschke’s theorem. Every finite-dimensional representation of a finite group is completely reducible: it decomposes as a direct sum of irreps.

So our $\mathcal V$ decomposes:

$$\mathcal V \;\cong\; V_{\alpha_1} \oplus V_{\alpha_2} \oplus \cdots$$

with each $V_{\alpha_i}$ an irrep of $P$.

What does this give us, qualitatively? Two things. First, $\mathcal V$ is built out of simple ingredients — irreducible pieces that cannot be split further while respecting symmetry. Second, for any given finite group $P$, there are only finitely many irreps up to equivalence. So the symmetry content of $\mathcal V$ is a list of pieces drawn from a small finite menu.

The same theorem applies to every eigenspace $E_\lambda$ separately, since each is itself a representation. So we now know each eigenspace is a sum of irreps drawn from the same finite menu — its symmetry content is also captured by a list.

Isotypic decomposition

Different copies of the same irrep can appear in $\mathcal V$, so it is clarifying to collect them. The isotypic decomposition groups equivalent irreps together:

$$\mathcal V \;\cong\; \bigoplus_\alpha V_\alpha \otimes M_\alpha.$$

The sum runs over irreps $V_\alpha$ of $P$ that appear in $\mathcal V$, and the multiplicity space $M_\alpha$ is a plain vector space (no $P$-action of its own) whose dimension counts how many copies of $V_\alpha$ appear. The notation $V_\alpha \otimes M_\alpha$ encodes ”$\dim M_\alpha$ copies of $V_\alpha$”: $P$ acts irreducibly on the $V_\alpha$ factor, trivially on $M_\alpha$.

The two factors play different roles. The $V_\alpha$ piece carries the symmetry content — how vectors there transform under $P$. The multiplicity space $M_\alpha$ carries the bookkeeping — how many copies of that symmetry type are present. The distinction matters in a moment.

Schur’s lemma constrains $\mathcal H$

Now we use $\mathcal H$‘s equivariance. Each isotypic component $V_\alpha \otimes M_\alpha$ is a $P$-invariant subspace, so $\mathcal H|_{\mathcal V}$ restricts to a $P$-equivariant operator on each. The decisive question: what can such an operator look like?

Schur’s lemma. Any $P$-equivariant linear map $V_\alpha \otimes M_\alpha \to V_\alpha \otimes M_\alpha$ has the form $\mathrm{id}_{V_\alpha} \otimes K_\alpha$ for some endomorphism $K_\alpha$ of the multiplicity space $M_\alpha$.

What does this tell us, qualitatively? $\mathcal H$ has no power to distinguish vectors within a single copy of an irrep. The whole copy is forced to behave the same way: $\mathcal H$ acts on it as the identity (times some scalar, set by where the copy sits in the multiplicity space). The only freedom $\mathcal H$ has is in how it mixes the different copies of an irrep among themselves. That mixing is encoded by $K_\alpha$ on $M_\alpha$.

The reason this is forced: irreps are by definition the smallest $P$-invariant subspaces; a $P$-equivariant map can rearrange copies of an irrep, but it cannot rearrange anything within a single copy without breaking equivariance.

Apply this to $\mathcal H|_{\mathcal V}$. On each isotypic component $V_\alpha \otimes M_\alpha$, $\mathcal H$ acts as $\mathrm{id}_{V_\alpha} \otimes K_\alpha$ for some self-adjoint operator $K_\alpha$ on $M_\alpha$ (self-adjointness inherited from $\mathcal H$). So $\mathcal H|_{\mathcal V}$ is a direct sum of self-adjoint operators on the multiplicity spaces, one block per irrep:

$$\mathcal H|_{\mathcal V} \;=\; \bigoplus_\alpha \bigl( \mathrm{id}_{V_\alpha} \otimes K_\alpha \bigr).$$

Aligning the two decompositions

We now have two direct-sum decompositions of $\mathcal V$:

• The frequency decomposition $\mathcal V = \bigoplus_\lambda E_\lambda$ from the spectral theorem — split by vibrational frequency.
• The isotypic decomposition $\mathcal V = \bigoplus_\alpha V_\alpha \otimes M_\alpha$ from rep theory — split by symmetry type.

There is a clear always-true relationship between them, plus a sharper generic statement on top. Let’s separate the two.

Always true: each eigenspace splits into whole irreps. This is just Maschke, applied one more time. Each eigenspace $E_\lambda$ is $P$-invariant (we proved this), so it is itself a representation of $P$. Apply Maschke directly to $E_\lambda$: it decomposes as a direct sum of irreps,

$$E_\lambda \;=\; \bigoplus_\alpha V_\alpha \otimes M_{\alpha, \lambda}.$$

The output of Maschke is by construction a sum of complete irreps — there is no “partial copy” possibility. So every irrep appearing inside $E_\lambda$ appears as a full copy, classified by its type $\alpha$, with $\dim M_{\alpha,\lambda}$ such copies (possibly $0$, if irrep $\alpha$ does not appear at frequency $\lambda$).

Comparing with the global isotypic decomposition $\mathcal V = \bigoplus_\alpha V_\alpha \otimes M_\alpha$: collecting the irrep-$\alpha$ pieces from each eigenspace recovers the global $\alpha$-component, with $M_\alpha = \bigoplus_\lambda M_{\alpha, \lambda}$. From the $\mathcal H$ side, this is exactly the spectral decomposition of $K_\alpha$ on $M_\alpha$ — its $\lambda$-eigenspace is $M_{\alpha,\lambda}$.

Generic: each eigenspace is one irrep. In general, an eigenspace $E_\lambda$ can contain pieces of multiple irrep types — if $K_\alpha$ and $K_\beta$ happen to share an eigenvalue $\lambda$, both contribute to $E_\lambda$. And $K_\alpha$ itself can have repeated eigenvalues, putting multiple copies of $V_\alpha$ into a single $E_\lambda$. Both are accidents: there is no symmetry reason for them, and a generic potential $V$ avoids them.

In the generic case — no accidents of either kind — the eigen decomposition coincides with the frequency decomposition itself, and each eigenspace $E_\lambda$ is exactly one copy of one irrep $V_\alpha$. Each vibrational frequency has a single “symmetry type,” and its eigenspace is $\dim V_\alpha$-dimensional.

What do we learn? Going back to the coarse questions we posed:

• How many distinct frequencies are there? Generically $\sum_\alpha n_\alpha$ — one for each eigenvalue of each $K_\alpha$. The total is set by the irrep multiplicities $n_\alpha := \dim M_\alpha$.
• How many independent modes at each frequency? $\dim V_\alpha$, where $V_\alpha$ is the irrep that frequency is “stamped with.” A 2-dimensional irrep forces pairs of degenerate modes; a 3-dimensional irrep forces triples.

Two different numbers in this story, easy to conflate:

• The multiplicity $n_\alpha := \dim M_\alpha$ — how many copies of $V_\alpha$ appear in $\mathcal V$. This counts the number of independent frequencies of symmetry type $\alpha$.
• The forced degeneracy $\dim V_\alpha$ — how many independent modes share each of those frequencies.

The list of irreps appearing in $\mathcal V$, together with their multiplicities $\{n_\alpha\}$, is the symmetry-only data of the spectrum: which irreducible blocks appear, at what dimension, with how many independent frequencies each. This list depends on $P$ and on the representation $\rho|_{\mathcal V}$ — both depending on the molecule and its symmetry, but neither depending on the potential.

The actual frequencies need $V$ (we still have to diagonalize each $K_\alpha$); the structural skeleton — the irrep at each frequency — does not.

Computing with characters

The decomposition $\mathcal V \cong \bigoplus_\alpha V_\alpha \otimes M_\alpha$ exists abstractly, but to actually pin down the multiplicities $\{n_\alpha\}$ we need a way to read them off the representation $\rho|_{\mathcal V}$ that doesn’t require us to explicitly diagonalize anything. Characters do this for us.

Given a finite-dimensional representation $\rho \colon P \to \mathrm{GL}(V)$, its character is the function

$$\chi_V \colon P \to \mathbb{R}, \qquad \chi_V(h) := \mathrm{tr}\,\rho(h).$$

The map $\rho \mapsto \chi_V$ seemingly throws almost all of our original representation away. $\chi_V$ is not a homomorphism — taking traces destroys multiplication — it is just a real-valued function on $P$, an element of the function space $\mathbb{R}^P$. The matrices are gone; only their traces remain.

What we trade up for is enormous simplification. $\mathbb{R}^P$ is a finite-dimensional vector space. The complicated category of representations gets replaced by a small linear-algebra problem.

For this trade to be worthwhile, the map had better be lossless — knowing $\chi_V$ should be enough to recover $V$ up to isomorphism. Amazingly, it is. This is the crucial theorem of character theory:

Characters determine representations. Two finite-dimensional representations of $P$ with the same character are isomorphic.

(There is no single agreed-upon name for this — Serre states it just as a corollary of the orthogonality relations below; it’s the load-bearing consequence of those relations.) Once we have this theorem, studying $V$ up to isomorphism is exactly the same as studying $\chi_V$ as an element of a small vector space. Everything we want to know — what irreps appear, with what multiplicities — is encoded in $\chi_V$, and is to be extracted by linear algebra.

The rest of this subsection is the structural setup that makes the linear algebra work cleanly.

$\chi_V$ is a class function. Trace is conjugation-invariant: $\chi_V(g h g^{-1}) = \chi_V(h)$. So $\chi_V$ doesn’t really take values on individual group elements — it takes values on conjugacy classes. The function space it actually lives in is the space of class functions

$$\mathcal{C}(P) \;:=\; \bigl\{\, f \colon P \to \mathbb{R} \;:\; f(g h g^{-1}) = f(h) \text{ for all } g, h \,\bigr\},$$

whose dimension equals the number of conjugacy classes of $P$. This is much smaller than $|P|$ — for $S_3$ ($|P| = 6$), $\dim \mathcal{C}(P) = 3$.

$\mathcal{C}(P)$ has a natural inner product, and irreducible characters are an orthonormal basis. The inner product is the group average,

$$\langle \chi, \chi' \rangle \;:=\; \frac{1}{|P|} \sum_{h \in P} \chi(h)\, \chi'(h) \;=\; \sum_{[h]} \frac{|[h]|}{|P|}\, \chi(h)\, \chi'(h),$$

(the second sum runs over conjugacy classes, legitimately because $\chi, \chi'$ depend only on the class). And:

Orthonormality of irreducible characters. The characters $\{\chi_\alpha\}_\alpha$ of the (finitely many) irreducible representations of $P$ form an orthonormal basis of $\mathcal{C}(P)$.

This is what does the work behind the scenes: linear independence of the $\chi_\alpha$ is what makes the multiplicities $n_\alpha$ in $\chi_V = \sum_\alpha n_\alpha \chi_\alpha$ uniquely determined — which is what makes “characters determine representations” true. Two consequences worth flagging: the number of irreps of $P$ equals the number of conjugacy classes of $P$ (both equal $\dim \mathcal{C}(P)$), and every class function expands uniquely against the basis $\{\chi_\alpha\}$.

Apply that last point to $\chi_V$ itself. If $V \cong \bigoplus_\alpha V_\alpha^{\oplus n_\alpha}$, additivity of trace gives $\chi_V = \sum_\alpha n_\alpha \chi_\alpha$, and orthonormality reads off the coefficients:

$$n_\alpha \;=\; \langle \chi_V,\, \chi_\alpha \rangle.$$

So the entire categorical question “what is $V$, up to isomorphism?” reduces to: compute the trace of $\rho(h)$ on $V$ for one $h$ in each conjugacy class — that’s all of $\chi_V$ — and take the inner product against each $\chi_\alpha$.

Character tables

A character table packages the irreducible characters of a fixed group $P$ into a single grid: rows indexed by irreps $V_\alpha$, columns indexed by conjugacy classes $[h]$, entries $\chi_\alpha([h])$.

By the orthonormality theorem, this table records everything needed for representation arithmetic over $P$. To find the multiplicities of irreps in any representation $V$, you need only

1. the character table of $P$ — a one-time lookup; and
2. the character $\chi_V$ of your representation — one number per conjugacy class.

Then $n_\alpha = \langle \chi_V, \chi_\alpha \rangle$ is a finite sum: pair the values $\chi_V([h])$ with the $\alpha$-row of the table, weight each term by class size, sum, divide by $|P|$.

A few features worth knowing for any character table. The first column (under $E$) is always the dimension of the irrep ($\chi_\alpha(E) = \dim V_\alpha$). The first row is always the trivial representation ($1$‘s everywhere — every element acts as the identity on $\mathbb{R}$). The number of rows equals the number of columns (both equal the number of conjugacy classes of $P$). And the squared dimensions of the irreps sum to the order of the group:

$$\sum_\alpha (\dim V_\alpha)^2 \;=\; |P|,$$

a useful sanity check.

Let’s now build the character tables for the two groups we will need: water’s and ammonia’s.

Water

Place water in the $xz$ plane with the oxygen at the origin and the $z$-axis bisecting the H–H line. Then $P$ has four elements:

• the identity $e$;
• the rotation $r$ by $\pi$ about the $z$-axis (sending each H to the other);
• the reflection $s$ in the molecular plane $xz$ (fixing all three atoms);
• the reflection $s'$ in the perpendicular plane $yz$ (sending each H to the other).

These satisfy $r^2 = s^2 = s'^2 = e$ and $r = s s'$; any two generate the third. The group is abelian, and abstractly it’s $\mathbb{Z}/2 \times \mathbb{Z}/2$ — the Klein four-group.

In an abelian group every conjugacy class is a singleton, so there are $4$ classes and hence $4$ irreps. The dimensions satisfy $\sum d_\alpha^2 = 4$, which forces $d_\alpha = 1$ for all $\alpha$ — every irrep is $1$-dimensional. A $1$-dimensional rep of the Klein four-group is just a homomorphism into $\{\pm 1\}$, so it amounts to picking signs for the two generators independently. Four sign-choice combinations, four irreps. Label them $V_{\eta_r, \eta_s}$ where $(\eta_r, \eta_s) \in \{\pm\}^2$ records the chosen signs:

• $V_{++}$: $r \mapsto +1$, $s \mapsto +1$ (the trivial rep).
• $V_{+-}$: $r \mapsto +1$, $s \mapsto -1$.
• $V_{-+}$: $r \mapsto -1$, $s \mapsto +1$.
• $V_{--}$: $r \mapsto -1$, $s \mapsto -1$.

Filling in the value at $s' = rs$ by multiplicativity (these are $1$-dim reps, so $\chi(rs) = \chi(r)\chi(s)$), the character table is:

	$e$	$r$	$s$	$s'$
$V_{++}$	$1$	$1$	$1$	$1$
$V_{+-}$	$1$	$1$	$-1$	$-1$
$V_{-+}$	$1$	$-1$	$1$	$-1$
$V_{--}$	$1$	$-1$	$-1$	$1$

The table is a $4 \times 4$ matrix of $\pm 1$‘s — orthogonality of rows is visible by inspection: any two distinct rows have equal counts of $+1$ and $-1$ in their pointwise product, summing to $0$. Each row’s squared norm is $4 = |P|$, normalizing to unit length under the $\frac{1}{|P|}\sum$ inner product.

Ammonia

Place ammonia with the nitrogen on the $z$-axis and the three H atoms forming an equilateral triangle in (a plane parallel to) the $xy$-plane. Then $P$ has six elements:

• the identity $e$;
• two rotations $r, r^2$ about the $z$-axis by $\pm 2\pi/3$ (cycling the three H atoms);
• three reflections $s_1, s_2, s_3$, one in each vertical plane through the $z$-axis and one of the H atoms.

These satisfy $r^3 = e$, $s_i^2 = e$, and $s_i r s_i = r^{-1}$ (each reflection inverts the rotation). This is the dihedral group $D_3$ — the symmetry group of an equilateral triangle — equivalently $S_3$, the symmetric group on the three H atoms (each $h \in P$ is determined by the permutation it induces on the three H labels).

Conjugacy classes: the rotations $r$ and $r^{-1} = r^2$ are conjugate to each other (any reflection conjugates one to the other), and the three reflections are all conjugate to each other (the rotations cycle them). So we get three classes:

• $\{e\}$ — $1$ element;
• $\{r, r^2\}$ — $2$ rotations;
• $\{s_1, s_2, s_3\}$ — $3$ reflections.

Three classes, hence three irreps. The dimensions satisfy $\sum d_\alpha^2 = 6$, with at least one $d_\alpha = 1$ (the trivial rep), and the unique solution in positive integers is $1 + 1 + 4 = 1^2 + 1^2 + 2^2$. So $S_3$ has two $1$-dimensional irreps and one $2$-dimensional irrep:

• The trivial rep $\mathbf{1}$: every element $\mapsto +1$.
• The sign rep $\mathrm{sgn}$: rotations $\mapsto +1$, reflections $\mapsto -1$. Under the identification $D_3 \cong S_3$, this sends each permutation to its sign.
• The standard rep $V_{\mathrm{std}}$: the action of $D_3$ on $\mathbb{R}^2$ as honest rotations and reflections of an equilateral triangle. (Equivalently: $S_3$ acts on $\mathbb{R}^3$ by permuting coordinates; $V_{\mathrm{std}}$ is the orthogonal complement of the diagonal $(1,1,1)$.)

To fill in the table we need traces. For the $1$-dim irreps the entries are just the sign assignments. For $V_{\mathrm{std}}$: the value at $e$ is $2$ (the dimension); a rotation by $2\pi/3$ in $\mathbb{R}^2$ has trace $2\cos(2\pi/3) = -1$; a reflection in $\mathbb{R}^2$ has trace $0$ (one $+1$ eigenvalue along the mirror, one $-1$ perpendicular). So:

	$\{e\}$	$\{r, r^2\}$	$\{s_1, s_2, s_3\}$
$\mathbf{1}$	$1$	$1$	$1$
$\mathrm{sgn}$	$1$	$1$	$-1$
$V_{\mathrm{std}}$	$2$	$-1$	$0$

Orthogonality is again checkable by inspection, now with the class-size weights: e.g. $\langle \chi_{\mathbf{1}}, \chi_{V_{\mathrm{std}}} \rangle = \tfrac16(1 \cdot 1 \cdot 2 + 2 \cdot 1 \cdot (-1) + 3 \cdot 1 \cdot 0) = 0$, and $\langle \chi_{V_{\mathrm{std}}}, \chi_{V_{\mathrm{std}}} \rangle = \tfrac16(1 \cdot 4 + 2 \cdot 1 + 3 \cdot 0) = 1$.

The standard rep $V_{\mathrm{std}}$ is the only $2$-dimensional one in either of our tables — it is the source of the forced doublet degeneracies we’ll find in ammonia’s vibrational spectrum.

Computing $\chi_{\mathrm{vib}}$

For our problem, $V = \mathcal V$. We get to $\chi_{\mathrm{vib}}$ by computing on the larger space $T_{\widehat e}\widehat Q$ and subtracting off the rigid-motion piece.

Total character $\chi_{\mathrm{total}}$ on $T_{\widehat e}\widehat Q = \bigoplus_i T_{x_i} X$. Each $\rho(h)$ is a block matrix in this decomposition. An atom $x_i$ that $h$ moves to a different atom $x_{\sigma_h(i)} \neq x_i$ contributes a zero diagonal block (its $T_{x_i}X$ entries land in $T_{x_{\sigma_h(i)}}X$). An atom that $h$ fixes contributes $\mathrm{tr}\bigl(h|_{T_{x_i}X}\bigr)$. So only fixed atoms contribute:

$$\chi_{\mathrm{total}}(h) \;=\; \sum_{i \,:\, \sigma_h(i) = i} \mathrm{tr}\bigl(h|_{T_{x_i}X}\bigr).$$

Rigid-motion character $\chi_N$ on $N \cong \mathfrak{g}/\mathfrak{p}$, with $P$ acting by the descended adjoint action. For $X = \mathbb R^3$ and a non-collinear molecule, $\mathfrak p = 0$ and $N \cong \mathfrak g \cong \mathbb R^3 \oplus \mathfrak{so}(3)$: three translations plus three rotations. $P$ acts on the translations by its inclusion $P \hookrightarrow O(3)$ (vector representation), and on the rotations by that same inclusion twisted by the determinant (axial-vector representation). So $\chi_N$ splits as

$$\chi_N(h) \;=\; \chi_v(h) \,+\, \det(h)\, \chi_v(h) \;=\; \bigl(1 + \det(h)\bigr)\, \chi_v(h),$$

where $\chi_v(h)$ is the trace of $h$ in the vector representation. This is $2\,\chi_v(h)$ on proper rotations and $0$ on improper ones.

Then

$$\chi_{\mathrm{vib}} \;=\; \chi_{\mathrm{total}} \,-\, \chi_N,$$

and $n_\alpha = \langle \chi_{\mathrm{vib}}, \chi_\alpha \rangle$.

Water

Coordinates as before — $z$-axis along the rotation axis $r$, molecule in the $xz$-plane, so $s$ is the molecular mirror and $s'$ is perpendicular. For each class:

• $e$ fixes all $3$ atoms; $\chi_v(e) = 3$, $\det = +1$. So $\chi_{\mathrm{total}}(e) = 9$, $\chi_N(e) = 6$.
• $r$ fixes only O (it swaps the H’s); $\chi_v(r) = -1$ (rotation by $\pi$ in $\mathbb R^3$ has trace $1 + 2\cos\pi = -1$), $\det = +1$. So $\chi_{\mathrm{total}}(r) = -1$, $\chi_N(r) = -2$.
• $s$ is the molecular plane and fixes all $3$ atoms; $\chi_v(s) = 1$ (trace of a reflection in $\mathbb R^3$), $\det = -1$. So $\chi_{\mathrm{total}}(s) = 3$, $\chi_N(s) = 0$.
• $s'$ fixes only O (swaps the H’s); $\chi_v(s') = 1$, $\det = -1$. So $\chi_{\mathrm{total}}(s') = 1$, $\chi_N(s') = 0$.

Tabulating:

	$e$	$r$	$s$	$s'$
$\chi_{\mathrm{total}}$	$9$	$-1$	$3$	$1$
$\chi_N$	$6$	$-2$	$0$	$0$
$\chi_{\mathrm{vib}}$	$3$	$1$	$3$	$1$

Inner-producting with each row of the character table (every class has size $1$, $|P| = 4$):

$$\begin{aligned} n_{V_{++}} &= \tfrac14\bigl(3 + 1 + 3 + 1\bigr) = 2,\\ n_{V_{+-}} &= \tfrac14\bigl(3 + 1 - 3 - 1\bigr) = 0,\\ n_{V_{-+}} &= \tfrac14\bigl(3 - 1 + 3 - 1\bigr) = 1,\\ n_{V_{--}} &= \tfrac14\bigl(3 - 1 - 3 + 1\bigr) = 0. \end{aligned}$$

So

$$\mathcal V_{\mathrm{water}} \;\cong\; 2\, V_{++} \oplus V_{-+}.$$

Total dimension $2 \cdot 1 + 1 \cdot 1 = 3$, matching $3n - 6 = 3$. ✓

What this tells us: water has $3$ vibrational modes, all stamped with $1$-dimensional irreps, so no forced degeneracies. The two $V_{++}$ modes are invariant under all of $P$ (these are the symmetric stretch and the bend); the $V_{-+}$ mode changes sign under both $r$ and $s'$ (the antisymmetric stretch). The two $V_{++}$ modes form a $2$-dimensional multiplicity space $M_{++}$, on which $K_{++}$ is a $2 \times 2$ self-adjoint operator — its two eigenvalues are the actual stretch and bend frequencies, set by the potential, with no symmetry obstruction to being whatever they are.

Ammonia

Three conjugacy classes — $\{e\}, \{r, r^2\}, \{s_1, s_2, s_3\}$.

• $e$: all $4$ atoms fixed; $\chi_v(e) = 3$, $\det = +1$. $\chi_{\mathrm{total}}(e) = 12$, $\chi_N(e) = 6$.
• $r$ (rotation by $2\pi/3$): only N fixed (the three H’s cycle); $\chi_v(r) = 1 + 2\cos(2\pi/3) = 0$, $\det = +1$. $\chi_{\mathrm{total}}(r) = 0$, $\chi_N(r) = 0$.
• $s_i$: N is fixed, plus the one H lying in this mirror plane; the other two H’s swap. So $2$ fixed atoms; $\chi_v(s_i) = 1$, $\det = -1$. $\chi_{\mathrm{total}}(s_i) = 2$, $\chi_N(s_i) = 0$.

	$\{e\}$	$\{r, r^2\}$	$\{s_1, s_2, s_3\}$
$\chi_{\mathrm{total}}$	$12$	$0$	$2$
$\chi_N$	$6$	$0$	$0$
$\chi_{\mathrm{vib}}$	$6$	$0$	$2$

Inner-producting (class sizes $1, 2, 3$; $|P| = 6$):

$$\begin{aligned} n_{\mathbf{1}} &= \tfrac16\bigl(1 \cdot 6 \cdot 1 \;+\; 2 \cdot 0 \cdot 1 \;+\; 3 \cdot 2 \cdot 1\bigr) = 2,\\ n_{\mathrm{sgn}} &= \tfrac16\bigl(1 \cdot 6 \cdot 1 \;+\; 2 \cdot 0 \cdot 1 \;+\; 3 \cdot 2 \cdot (-1)\bigr) = 0,\\ n_{V_{\mathrm{std}}} &= \tfrac16\bigl(1 \cdot 6 \cdot 2 \;+\; 2 \cdot 0 \cdot (-1) \;+\; 3 \cdot 2 \cdot 0\bigr) = 2. \end{aligned}$$

So

$$\mathcal V_{\mathrm{ammonia}} \;\cong\; 2\, \mathbf{1} \oplus 2\, V_{\mathrm{std}}.$$

Total dimension $2 \cdot 1 + 2 \cdot 2 = 6$, matching $3n - 6 = 6$. ✓

Now we see the forced doublets explicitly: ammonia has $6$ vibrational modes, organized as $2$ singlets (copies of the trivial rep $\mathbf{1}$) plus $2$ doublets (copies of $V_{\mathrm{std}}$). The $\mathbf{1}$-block $K_{\mathbf{1}}$ is $2 \times 2$ on $M_{\mathbf{1}}$, contributing two singlet frequencies; the $\mathrm{std}$-block $K_{\mathrm{std}}$ is $2 \times 2$ on $M_{\mathrm{std}}$, and each of its eigenvalues comes with multiplicity $2$ in the full spectrum, because $V_{\mathrm{std}}$ is $2$-dimensional. So generically ammonia’s spectrum has $4$ distinct frequencies: two nondegenerate (the two trivial-rep copies) plus two doubly degenerate (the two standard-rep copies). Physically these are commonly described as a symmetric stretch and an “umbrella” inversion (the singlets) plus an asymmetric stretch pair and an asymmetric bend pair (the doublets).

What this computes — and what it doesn’t

What the character procedure gives us is the list $\{n_\alpha\}$, which by Schur tells us exactly how the spectrum decomposes by irrep — how many singlets, how many doublets, how many triplets, with all forced degeneracies accounted for. We learn the structural skeleton of the spectrum from $P$ and the molecule alone, before any eigenvalue problem is solved.

What we don’t get is the numerical values of the frequencies. Each $K_\alpha$ is a small self-adjoint operator on the multiplicity space $M_\alpha$, and to compute its eigenvalues we need $V$. Representation theory gives the structure; the dynamics gives the numbers.

This is the punchline. The character-table procedure that chemists use is doing exactly the symmetry-only part of the analysis — neither more nor less than what character orthogonality lets representation theory see. The remaining numerical eigenvalue problem inside each $\alpha$-block is a separate, smaller calculation, doable molecule-by-molecule once a potential is chosen.

Epilogue: What the generality bought us

Working with an abstract $(X, g)$ throughout, rather than narrowing to $\mathbb R^3$ from the start, was free — the derivation never used anything about $X$ that wasn’t packaged into “Riemannian manifold with isometry group.” As a payoff, we get a few structural facts that aren’t obvious from the chemistry-textbook treatment.

Curvature is invisible to the harmonic machinery. The Riemann curvature of $\widehat{\mathbf g}$ made a brief appearance in the linearization (in the $R(S, T)\, T$ term) and then vanished, because $T = 0$ at the equilibrium. The whole construction — the operator $\mathcal H$, its self-adjointness, the rep-theoretic decomposition of its spectrum — went through without ever using the curvature of $(X, g)$. Curvature first enters at cubic order, in the anharmonic corrections.

(The numerical eigenvalues of $\mathcal H$ do still depend on the ambient space, since the potential $V$ is typically built from pairwise geodesic distances, which differ between $\mathbb R^3$, $S^3$, $\mathbb H^3$, and so on. What’s curvature-independent is the form of the harmonic theory — same construction, same self-adjointness, same Schur structure — not the values of the frequencies.)

The vibrational dimension is $n \dim X - \dim G$. For a non-collinear molecule the dimension of the vibrational subspace $\mathcal V$ is the total degrees of freedom minus the dimension of the ambient isometry group:

$$\dim \mathcal V \;=\; n \dim X \;-\; \dim G.$$

For a molecule of $n$ atoms in $\mathbb R^3$ ($\dim G = 6$) this is the chemist’s familiar $3n - 6$. On $S^3$ and $\mathbb H^3$ it is also $3n - 6$ ($\dim G = 6$ in both cases). On $\mathrm{Nil}$ ($\dim G = 4$) it is $3n - 4$, and on $\mathrm{Sol}$ ($\dim G = 3$) it is $3n - 3$ — different ambient geometries support different numbers of “rigid motions” of a generic molecule, and the vibrational dimension absorbs the difference.

Mode organization is rep theory of $P$ on $\mathcal{V}$. The way the spectrum is organized into blocks — which irreps appear, with what multiplicities, with what forced degeneracies — is determined entirely by the representation $\rho|_{\mathcal V}$ of the symmetry group $P$ on the vibrational subspace. This depends on the molecule, but not on the potential, and not on whether the ambient space is curved or flat (but it does depend on the size of the isometry group). Same character tables, same forced degeneracies, regardless of $(X, g)$.