Morse Deformation, Gradient Flow, and Semiclassical Tunneling

On a compact Riemannian manifold, conjugating the de Rham differential by $e^{tf}$ leaves its cohomology unchanged for every finite $t$, yet makes the large-$t$ Hamiltonian concentrate near the critical points of the Morse function $f$. Each critical point contributes one local low-energy state in the degree equal to its Morse index. Those local states are not generally exact vacua: exponentially small tunneling along signed gradient-flow trajectories reconstructs the Morse differential and removes the combinations that do not represent cohomology.

Required background. Q-cohomology and Hodge decomposition supplies the de Rham Hilbert complex, while Laplace’s method and steepest descent supplies the large-parameter expansion. Helpful background. Zero modes, collective coordinates, and moduli measures develops the general semiclassical treatment; this page uses only the finite-dimensional supersymmetric instance.

The Witten-deformed complex

Let $M$ be a smooth, compact, oriented Riemannian manifold without boundary, and let $f:M\to\mathbb R$ be a Morse function: every critical point is isolated and has nonsingular Hessian. For $t\ge0$, define

$$d_t=e^{-tf}d\,e^{tf}=d+t\,df\wedge, \qquad d_t^\dagger=e^{tf}\delta\,e^{-tf} =\delta+t\,\iota_{\nabla f}.$$

Because $d_t$ is conjugate to $d$,

$$d_t^2=0, \qquad H_t^\bullet(M)\cong H_{\mathrm{dR}}^\bullet(M).$$

The isomorphism is induced by multiplication by $e^{-tf}$. Compactness matters here: for finite $t$, both $e^{tf}$ and $e^{-tf}$ are bounded and preserve the relevant Sobolev domains. The supersymmetric Hamiltonian is

$$H_t=\frac12\Delta_t, \qquad \Delta_t=d_td_t^\dagger+d_t^\dagger d_t.$$

In a local orthonormal frame, let $\varepsilon^i$ denote exterior multiplication by the coframe and $\iota^j$ contraction by the dual frame. Direct expansion gives

$$\Delta_t =\Delta_{\mathrm{dR}} +t^2\lvert df\rvert^2 +t\,(\nabla_i\nabla_j f)[\varepsilon^i,\iota^j].$$

The three terms have distinct jobs: the ordinary form Laplacian controls kinetic energy, $t^2|df|^2$ confines low-energy states near $df=0$, and the Hessian term assigns their form degree. This is the operator at the center of Witten’s construction Witten 1982, §2, pp. 665–667.

One local state per critical point

Let $p$ be a critical point. Choose normal coordinates $y^i$ that diagonalize the Hessian,

$$f(y)=f(p)+\frac12\sum_{i=1}^{n}\lambda_i(y^i)^2+O(|y|^3), \qquad \lambda_i\ne0.$$

To leading order, $\Delta_t$ is a sum of bosonic oscillators and commuting fermionic two-state systems. Its unique local zero-energy state is

$$\Phi_p(y)\propto \exp\!\left[-\frac t2\sum_i|\lambda_i|(y^i)^2\right] \bigwedge_{\lambda_i<0}\mathrm dy^i.$$

The Gaussian has width $O(t^{-1/2})$. The wedge contains one factor for each negative Hessian eigenvalue, so its degree is

$$\deg\Phi_p=\operatorname{ind}_f(p),$$

the Morse index of $p$. All other local oscillator states have energies of order $t$. Thus, if $C^k$ is spanned by critical points of index $k$, the large-$t$ low-energy space has dimension $\dim C^k=m_k$, the number of such critical points.

This is a local asymptotic statement, not yet a count of exact zero modes. It immediately gives the weak Morse inequalities $b_k\le m_k$, because the exact number of harmonic $k$-forms is the Betti number $b_k$, while no other state can remain near zero as $t\to\infty$ Witten 1982, pp. 666–668.

Gradient flow from the Euclidean action

The bosonic Euclidean action governing a trajectory $x(\tau)$ is, up to the same normalization used in $H_t$,

$$S_E[x] =\frac12\int_{-\infty}^{+\infty} \left(|\dot x|^2+t^2|\nabla f|^2\right)\mathrm d\tau.$$

Completing the square gives either orientation,

$$S_E =\frac12\int|\dot x\pm t\nabla f|^2\mathrm d\tau \mp t\,[f(x(+\infty))-f(x(-\infty))].$$

Therefore a path between critical points obeys

$$S_E\ge t\,|f(q)-f(p)|,$$

with equality on a gradient trajectory

$$\dot x=\mp t\nabla f.$$

After rescaling $\tau$, these are precisely the upward or downward Morse flows. The exponential factor in a tunneling amplitude is $e^{-t|f(q)-f(p)|}$ Witten 1982, pp. 671–672.

Fermion zero modes impose the selection rule, but their number requires a transversality hypothesis. For a trajectory from $q$ to $p$ with relative Morse index

$$r=\operatorname{ind}_f(q)-\operatorname{ind}_f(p),$$

the linearized trajectory operator has Fredholm index $r$. If $f$ and the metric are Morse–Smale, the operator is surjective, so its kernel has dimension $r$. One kernel direction is translation of the trajectory in $\tau$; therefore the unparametrized flow space has dimension $r-1$. The trajectories that enter the degree-one differential are isolated after quotienting by translation, hence have $r=1$ and exactly one fermionic zero mode for the $d_t$ insertion to absorb. This is the selection rule in Witten 1982, p. 672.

Tunneling reconstructs the differential

Under these Morse–Smale hypotheses, choose orientations of the unstable manifolds. For a critical point $p$ of index $k$ and $q$ of index $k+1$, let $n(q,p)$ be the signed count of isolated downward flows from $q$ to $p$. In a normalized local basis,

$$\langle q|d_t|p\rangle \sim n(q,p)e^{-t[f(q)-f(p)]}.$$

Nonzero bosonic and fermionic fluctuation determinants cancel in magnitude. The translational bosonic zero mode cancels the normalization of the single fermionic zero mode; the remaining sign is the orientation sign of the trajectory. Rescaling the basis vectors by their critical values removes the displayed exponent and leaves the Morse cochain differential

$$\delta|p\rangle =\sum_{\operatorname{ind}(q)=k+1}n(q,p)|q\rangle.$$

The relation $\delta^2=0$ is not a cancellation one may assume trajectory by trajectory. The one-dimensional moduli spaces of flows between index difference two compactify by adding broken trajectories; their oriented boundary points cancel in pairs. Equivalently, $\delta$ is the large-$t$ restriction of $d_t$, whose square is exactly zero. Witten derives the signed trajectory operator and its semiclassical matrix elements in Witten 1982, pp. 668–675.

The exact ground states are the cohomology of this finite complex:

$$H^k(C^\bullet,\delta)\cong H^k_{\mathrm{dR}}(M).$$

Individual critical points are therefore approximate vacua. Only cohomology classes of their signed combinations survive at exactly zero energy.

Example: two trajectories on a circle

Take $M=S^1$ and $f(\theta)=\cos\theta$. There is one maximum at $\theta=0$ with Morse index $1$ and one minimum at $\theta=\pi$ with index $0$. The large-$t$ approximation produces one localized one-form near the maximum and one localized zero-form near the minimum.

There are two downward gradient trajectories from the maximum to the minimum, one around each side of the circle. Their actions are equal, $S_E=2t$, but their orientation signs are opposite. Consequently their contributions cancel:

$$n(\text{maximum},\text{minimum})=(+1)+(-1)=0.$$

The Morse differential vanishes. Its cohomology has one generator in degree zero and one in degree one, reproducing

$$b_0(S^1)=b_1(S^1)=1.$$

This example shows why counting critical points is insufficient and why determinant signs cannot be discarded. Each tunneling path is nonzero; the signed sum is zero.

A deformation that deliberately fails

Compactness cannot be omitted from the conjugation argument. On $M=\mathbb R$, start with the $L^2$ de Rham complex. At $t=0$, the free Laplacian has continuous spectrum down to zero and neither the constant zero-form nor the constant one-form is square-integrable, so there is no harmonic $L^2$ state.

Now take $f(x)=x^2/2$ and $t>0$:

$$d_t=d+t x\,\mathrm dx\wedge.$$

The degree-zero equation $d_t\psi=0$ has the normalized solution

$$\psi_t(x)=\left(\frac{t}{\pi}\right)^{1/4}e^{-t x^2/2}.$$

A normalizable supersymmetric ground state has appeared. There is no contradiction: multiplication by $e^{tf}=e^{tx^2/2}$ is unbounded on $L^2(\mathbb R)$, $d_t$ is not related to $d$ by a bounded invertible map of the physical Hilbert complexes, and the spectral gap closes as $t\to0$. At the endpoint, the operator is non-Fredholm and the formal index-invariance proof has lost its hypotheses.

This is the model to remember before using a localization-style deformation in infinite volume: algebraic conjugacy of differential expressions does not preserve $L^2$ cohomology by itself.

Limits of the semiclassical construction

• If $f$ is Morse–Bott rather than Morse, a critical submanifold contributes its own differential-form complex, twisted when the negative normal bundle is not orientable.
• Without Morse–Smale transversality, flow spaces need a perturbation or a more careful virtual construction before a signed count is defined.
• On a noncompact manifold, flows can escape to infinity and multiplication by $e^{tf}$ can change domains and normalizability.
• With a boundary, the gradient flow and the supercharge domain require compatible boundary conditions; extra boundary trajectories may contribute.
• The semiclassical expansion controls the large-$t$ regime. It does not turn every approximate local zero mode into an exact state.

Exercises

Instanton action. Complete the square in $S_E$ for a downward flow from a critical point $q$ to $p$ with $f(q)>f(p)$, and identify the action of a saturating trajectory.

Solution

Use

$$S_E=\frac12\int|\dot x+t\nabla f|^2\mathrm d\tau +t[f(q)-f(p)].$$

The first term vanishes for $\dot x=-t\nabla f$, so the minimum action is $t[f(q)-f(p)]$. Its contribution is proportional to $e^{-t[f(q)-f(p)]}$, before the basis rescaling used to define the integer Morse differential.

Signed trajectories, not an unsigned count. Let $p$ have Morse index zero. Suppose two index-one critical points $q_1$ and $q_2$ have signed trajectory counts

$$n(q_1,p)=(+1)+(-1), \qquad n(q_2,p)=+1.$$

Compute $\delta|p\rangle$ and explain what would go wrong if the trajectories were counted without signs.

Solution

The two trajectories to $q_1$ cancel, whereas the trajectory to $q_2$ does not. Therefore

$$\delta|p\rangle=|q_2\rangle.$$

An unsigned count would incorrectly give coefficient two for $|q_1\rangle$ and would generally destroy the oriented-boundary cancellation behind $\delta^2=0$. The signs are fixed by orientations of determinant lines, or equivalently of the unstable manifolds; they are not optional phases that can be discarded after computing the instanton action.

References

• Witten, Edward. “Supersymmetry and Morse Theory.” Journal of Differential Geometry 17, no. 4 (1982): 661–692. doi:10.4310/jdg/1214437492.