Physics for a Mathematician 02-IV - Supersymmetric Quantum Mechanics from a Morse Function

IV. Putting it all Together

Supersymmetric Quantum Mechanics from a Morse Function

IV.1: Hodge Theory basics

We are almost ready to jump into Witten's construction. To begin, we describe some basics of Hodge Theory.

Suppose $V$ is a vector space (over $\mathbb{R}$, for now) of finite dimension $n$ with an inner product, which we think of as a positive-definite symmetric $2$-tensor $g$ on $V$ (so we will write $g(v,w)$ instead of $\langle v,w \rangle$. Let $e_1, \ldots, e_n$ be an orthonormal basis, i.e. such that $g(e_i,e_j) = \delta_{ij}$. Let $\phi_1, \ldots, \phi_n$ be the dual basis for $V^*$, the space of linear functionals on $V$: in other words $\phi_i(v) := g(e_i,v)$. In particular, there is an identification $V \cong V^*$ by identifying $e_i$ and $\phi_i$.

Exercise: Show that this identification $V \cong V^*$ is canonical, as it is just the identification of a vector $v \in V$ with the functional $\phi_v := g(v, \cdot) \in V^*$ with $\phi_v(w) = g(v,w)$.

Let $\Lambda^k(V)$ be the vector space of alternating $k$-tensors on $V$, which comes with a natural basis of elements $$\phi_I := \phi_{i_1} \wedge \cdots \wedge \phi_{i_k},$$ where $I = (i_1,\ldots, i_k)$ is a strictly increasing sequence of indices $1 \leq i_1 < \cdots < i_k \leq n$. We define an inner product on $\Lambda^k(V)$, also denoted by $g$, by declaring the elements $\phi_I$ to be orthonormal. A priori, this all depends upon our initial choice of orthonormal basis $e_1, \ldots, e_n$, but actually:

Exercise: Show that the inner product $g$ on each $\Lambda^k(V)$ is independent of the original choice of orthonormal basis $e_1, \ldots, e_n$.

Hence, since $\Lambda^n(V)$ is $1$-dimensional, every element is of the form $c\omega$, with $\omega = \phi_{1,2,\ldots,n}$. We have that $\|\omega\| = 1$, so $\omega$ is determined up to multiplication by $\pm 1$, and this is fixed by an orientation of $V$ (corresponding to whether $(e_1,\ldots,e_n)$ is an oriented frame. From now on, we suppose that we are given an orientation of $V$ so that we may talk about $\omega$ without reference to a basis.

Definition: The Hodge star operator $\star \colon \Lambda^k(V) \rightarrow \Lambda^{n-k}(V)$ is defined by the property that for for all $\alpha \in \Lambda^{k}(V)$, we have $$\alpha \wedge \star \beta = g(\alpha,\beta)\omega.$$

Exercise: This defines a linear isomorphism with the property that $\star\star = (-1)^{k(n-k)}.$

Example: We have

• $\star 1 = \omega$
• $\star \phi_i = (-1)^{i-1}\phi_1 \wedge \cdots \wedge \phi_{i-1} \wedge \widehat{\phi_i} \wedge \phi_{i+1} \wedge \cdots \wedge \phi_n$
• In general, $\star \phi_I = \pm \phi_J$ where $I \sqcup J = \{1,\ldots,n\}$ forms a partition of the indices $1$ through $n$. The sign is the same one appearing in the equation $\phi_I \wedge \phi_J = \pm \phi_{1,2,\ldots,n}$.

Suppose now that $(M,g)$ is a Riemannian manifold. Each point $p \in M$ comes with the inner product space $(T_pM, g_p)$, and hence a Hodge star operator at each point. Hence, applying the Hodge star operator pointwise, we obtain an operator on differential forms $$\star \colon \Omega^k(M) \rightarrow \Omega^{n-k}(M).$$ In particular, we obtain a volume form $$\Omega := \star 1,$$ i.e. a top-dimensional non-vanishing form. We also have the exterior differential $d \colon \Omega^k(M) \rightarrow \Omega^{k+1}(M)$. We may construct a linear map in the opposite direction as follows:

Definition: The codifferential $d^* \colon \Omega^{k+1}(M) \rightarrow \Omega^k(M)$ is defined by $$d^*(\omega) = (-1)^{nk+1}\star d \star(\omega).$$

Exercise: The codifferential is more naturally defined as an $L^2$-adjoint for $d$ (explaining the notation), meaning that it satisfies $$\int_M \langle d\alpha, \beta \rangle \Omega = \int_M \langle \alpha, d^*\beta \rangle \Omega$$ for all $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^{k+1}(M)$. Show that this is indeed an equivalent definition.

Example: Suppose we take $M = \mathbb{R}^n$ with its standard orientation and metric. Consider a $1$-form $\eta = \sum f_i dx_i$. Then $$\begin{align*}d^*\eta &= -\star d \star \sum f_i dx_i \\ &= -\sum \star d((-1)^{i-1}f_i dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge d_n) \\ &= -\sum \star \frac{\partial f_i}{\partial x_i}\Omega \\ &= -\sum \frac{\partial f_i}{\partial x_i}\end{align*}$$

We see from this example that, up to isomorphism, $d^*$ applied to a $1$-form yields a notion of divergence. (This is $\star$-dual to applying $d$ to $(n-1)$-forms, which is another way to think of the divergence.) So this is a very natural operation. The duality between $d$ and $d^*$ tells us that any result holding for $d$ should have a dual statement holding for $d^*$. For example, we see that $(d^*)^2 = 0$.

Having $d$ and $d^*$ as dual objects, there is a very natural algebraic operation to put both of them together. We may form the vector space $$\Omega^*(M) := \bigoplus \Omega^k(M),$$ so that $d$ and $d^*$ both act on this vector space. In fact, this is a super vector space, with parity given by separating the even and odd differential forms, and $d$ and $d^*$ are each odd operators. This allows us to form the odd operator $Q = d+d^*$, which puts $d$ and $d^*$ on a symmetric footing. It satisfies $$\Delta := Q^2 = (d+d^*)^2 = d^2 + dd^*+d^*d +(d^*)^2 = dd^* + d^*d.$$

Definition: The operator $\Delta$ is called the (geometer's) Laplacian.

Example: For $f \in \Omega^0(M)$, we have $d^*f = 0$ (since there are nonzero $(-1)$-forms), and so $$\Delta f = d^*df = d^* \sum \frac{\partial f}{\partial x_i}dx_i = - \sum \frac{\partial^2 f}{\partial x_i^2}.$$ (The analyst's Laplacian comes with the opposite sign, which is more standard for a course in analysis.)

Definition: An element in the kernel of the Laplacian $\Delta$ is called a harmonic form. The space of harmonic $k$-forms is denoted $\mathcal{H}^k(M)$.

Remark: It is for this reason that in the last post, I used $\mathcal{V}$ to denote our quantum mechanical Hilbert space. I will continue with this notation later so that there is no confusion with harmonic forms.

Definition: A differential form $\eta \in \Omega^{k}(M)$ is called closed if $d\eta = 0$ and exact if we may write $\eta = d\alpha$ for some $\alpha \in \Omega^{k-1}(M)$. Dually, it is called coclosed if $d^*\eta = 0$ and coexact if we may write $\eta = d^*\beta$ for some $\beta \in \Omega^{k+1}(M)$.

Proposition: On a closed Riemannian manifold $(M,g)$, a differential $k$-form is harmonic if and only if it is both closed and coclosed.

Proof: Consider the $L^2$-inner product on differential $k$-forms given by $$\langle \alpha, \beta \rangle = \int_M g(\alpha,\beta) \Omega.$$ Then if $h \in \Omega^k(M)$ is harmonic, we have $\Delta h = 0$, and hence $$0 = \langle \Delta h, h \rangle = \langle dd^*h, h \rangle + \langle d^*dh,h \rangle = \langle dh,dh \rangle + \langle d^*h, d^*h \rangle.$$ Both terms on the right are non-negative, so they must both be zero, and so any harmonic form is both closed and coclosed. Conversely, if $h$ is closed and coclosed, then $\Delta h = (dd^*+d^*d) h = 0$ is clear.

The key result of Hodge theory is the following.

Hodge Decomposition Theorem: Suppose $(M,g)$ is a closed Riemannian manifold (NB: It is important that the manifold is closed!) Then there is an $L^2$-orthogonal decomposition $$\Omega^k(M) = \mathcal{H}^k(M) \oplus d(\Omega^{k-1}(M)) \oplus d^*(\Omega^{k+1}(M)).$$ In other words, every differential $k$-form $\omega$ can be written uniquely as $$\omega = h + \alpha + \beta$$ where $h$ is harmonic, $\alpha$ is exact, and $\beta$ is coexact, and these forms automatically satisfy  $$\int_M g(h,\alpha)\Omega = \int_M g(h,\beta)\Omega = \int_M g(\alpha,\beta)\Omega = 0.$$

Exercise: Check the $L^2$-orthogonality property of the components.

We will not prove the Hodge Decomposition Theorem, but we will work a little harder to see how it fits into the story of de Rham cohomology.

Corollary: If $\alpha$ is coclosed and exact, then it is $0$. Dually, if $\beta$ is closed and coexact, then it is $0$.

Proof: Since $\alpha$ is exact, it is certainly closed, and so $\alpha$ is closed and coclosed, hence harmonic. But the harmonic and exact forms are $L^2$-orthogonal, so $\alpha = 0$. The dual proof gives the result for $\beta$.

Corollary: The closed $k$-forms are given by $\mathcal{H}^k(M) \oplus d(\Omega^{k-1}(M))$.

Proof: This is immediate from the previous corollary.

Recall now the following:

Definition: The $k$th de Rham cohomology $H^k(M;\mathbb{R})$ of a manifold $M$ is the vector space quotient $$H_{\mathrm{dR}}^k(M;\mathbb{R}) := \{\mathrm{closed}~k\mathrm{-forms}\}/\{\mathrm{exact}~k\mathrm{-forms}\}.$$

Theorem/Corollary: For each de Rham cohomology class on a Riemannian manifold $(M,g)$, there is a unique harmonic form representing that cohomology class, i.e. there is a canonical isomorphism $$H_{\mathrm{dR}}^k(M;\mathbb{R}) \cong \mathcal{H}^k(M,g).$$

Remark: Harmonic forms are highly dependent on the metric. The right-hand side of the association is meaningless without the choice of a metric, whereas the left-hand side is purely topological, without reference to a metric. This represents an interesting analytic way of thinking of de Rham cohomology.

Remark: One could repeat the construction of de Rham cohomology but with coclosed and coexact forms. Again, we obtain an isomorphism with harmonic forms, so these dual complexes have canonically isomorphic cohomologies.

As one final step, we note that at every stage, we can extend the underlying field from $\mathbb{R}$ to $\mathbb{C}$. In other words, every differential form is now of the form $\sum f_I dx_I$ in local coordinates, but where $f_I \colon M \rightarrow \mathbb{C}$. Nothing in the story discussed thus far changes, so we work over $\mathbb{C}$ from now on without further comment.

IV.2: Supersymmetric Quantum Mechanical Systems from a Morse function

Let us now delve into Witten's construction. We begin with the input of a Morse function $f \colon M \rightarrow \mathbb{R}$ on a smooth manifold $M$. We also fix a Riemannian metric $g$ on the manifold $M$. For each $t \geq 0$, we construct a supersymmetric quantum mechanical system.

To start, we will construct the version with $t = 0$. For this, we need a super Hilbert space. Let $\Omega^p(M)$ be the space of ($C^{\infty}$, complex-valued) differential $p$-forms on $M$. Our super Hilbert space $\mathcal{V}$ has $$\mathcal{V}_0 := \bigoplus_{p~\mathrm{even}} \Omega^p(M),~~~~~\mathcal{V}_1 = \bigoplus_{p~\mathrm{odd}} \Omega^p(M),$$ and the inner product is the one coming from Hodge theory. We then have the supersymmetric quantum mechanical system given by $$Q_1 = d+d^*,~~~~~Q_2 = i(d-d^*),~~~~~H = \Delta_g = dd^* + d^*d.$$ One checks that this indeed satisfies the desired properties! In particular, standard Hodge theory applies, and we have that the kernel of $H$ consists of the (super vector space of) harmonic forms, $$\mathrm{ker}(H) = \mathcal{H}^*(M),$$ with grading given by the parity of the degree of the form. As noted by the section about Hodge theory, this is finite-dimensional, just given by the (complex-valued) de Rham cohomology.

For $t > 0$, we deform this picture by setting $$d_t = e^{-tf}de^{tf}~~~~~d_t^* = e^{tf}d^*e^{-tf}.$$ These are again $L^2$-adjoint operators (with respect to the $L^2$-inner product on differential forms induced by $g$), and replacing our operators by the corresponding expressions $$Q_{1,t} = d_t+d_t^*,~~~~~Q_{2,t} = i(d_t-d_t^*),~~~~~H_{t} = \Delta_t = d_td_t^* + d_t^*d_t.$$  

Much of what holds for standard Hodge theory also holds in this general setting, including the Hodge decomposition. One finds that the kernel of $H_t$, which we'll call the twisted harmonic forms  $$\mathcal{H}_t := \mathrm{ker}(H_{t}),$$  is naturally identified with the cohomology of the complex $(\Omega^*(M), d_t)$. But this is isomorphic to the standard de Rham cohomology since the map $\alpha \mapsto e^{tf}\alpha$ yields an isomorphism of the underlying cochain complexes, i.e. the following is a commutative diagram such that the vertical arrows are isomorphisms: $$\begin{matrix} \cdots & \xrightarrow{d_t} & \Omega^{k-1}(M) & \xrightarrow{d_t} & \Omega^k(M) & \xrightarrow{d_t} & \Omega^{k+1}(M) & \xrightarrow{d_t} & \cdots \\  & & \left.{e^{tf}}\right\downarrow & & \left.{e^{tf}}\right\downarrow & & \left.{e^{tf}}\right\downarrow & & \\ \cdots & \xrightarrow[d]{} & \Omega^{k-1}(M) & \xrightarrow[d]{} & \Omega^k(M) & \xrightarrow[d]{} & \Omega^{k+1}(M) & \xrightarrow[d]{} & \cdots \end{matrix}$$

We have our supersymmetric quantum mechanical system, and we aim now to study the eigenstates of $H_{t}$. This includes the twisted harmonic forms $\mathcal{H}_t$ just described, which form the vacuum states, as they have the lowest possible energy.

The key is to understand what $H_{t}$ looks like in local coordinates.

Theorem / Tricky Exercise: Suppose $x_1, \ldots, x_n$ are local coordinates on $M$. Fix an orthonormal frame $(e_1,\ldots,e_n)$ on $M$ in this coordinate chart. Let $a^i \colon \Omega^{k+1}(M) \rightarrow \Omega^k(M)$ be the operator given by $$a^i\eta = \eta(e_i, \cdot, \ldots, \cdot)$$ and let $(a^i)^*$ be the dual operator with respect to the metric. Then with respect to the local coordinate and frame, we may write $$H_{t} = \Delta + t^2A + tB,$$ where $A$ and $B$ are described as follows:

• $A$ is multiplication by the function $g(df,df) = \|df\|^2$, with respect to the metric on $\Omega^1(M)$
•  $B$ is the operator $\sum_{i,j}\mathrm{Hess}(f)(e_i,e_j) \cdot [(a^i)^*,a^j]$.

Remarks:

• Although on a general manifold, the Hessian is only well-defined at the critical points, on a Riemannian manifold, one may form the Hessian more generally, and it will be a $2$-tensor, explaining how $e_i$ and $e_j$ appear as inputs for the Hessian in the above expression. This matches with the standard description as a matrix of second partial derivatives in local coordinates at the critical points (c.f. the first blog post in this series).
• Recall that the commutator of two operators is $[\mathcal{O}_1,\mathcal{O}_2] = \mathcal{O}_1 \mathcal{O}_2 - \mathcal{O}_2\mathcal{O}_1$.

Notice, therefore, that studying the time-independent equation $H_t \eta = E\eta$ essentially amounts to studying the standard Schrödinger equation, but where the potential energy is taken instead to be the operator $t^2A + tB$. For $t$ large, the term $tB$ is small relative to $t^2A$, so we may think of the potential energy as approximately given by $t^2A$, which is simply multiplication by a function, and so we enter the realm of differential equations if this approximation is valid.

Physically, for large $t$, the term $t^2A$ is very big unless we are near a critical point, where $A = 0$. In other words, if we have an eigenstate, we expect it to be very small everywhere except for the critical points. It therefore makes sense, for large $t$, to study the eigenstates near the critical points.

It is now a standard fact from Morse theory and Riemannian geometry that, around each critical point $p$, we may find local coordinates $x_1,\ldots,x_n$ such that the point $p$ is located at the origin and $$g_{ij}(x) = \delta_{ij} + o(\|x\|^2)$$ (i.e. $g_{ij}(0) = \delta_{ij}$ and $\frac{\partial g_{ij}}{\partial x_k} = 0$), and in addition, $$f(x) = f(p) + \frac{1}{2}\sum \lambda_i x_i^2 + o(\|x\|^3).$$ Then we may use the approximately orthonormal frame $\left(\frac{\partial}{\partial x_1}, \cdots, \frac{\partial}{\partial x_n}\right)$ to define $a^i$ and $(a^i)^*$, and one finds $$H_t \approx \overline{H_t} := \sum_{i=1}^{n} \left(-\frac{\partial^2}{\partial x_i^2} + t^2\lambda_i^2 x_i^2 +t\lambda_i \cdot [(a^i)^*,a^i]\right),$$ where the approximation is up to order $o(\|x\|^3)$. By perturbation theory, the details of which we will ignore, one may approximate eigenstates and eigenvalues of $H_t$ (near the critical point $p$) by studying the eigenstates and eigenvalues of $\overline{H_t}$. We will ignore the inner workings of this detail, but will state a precise theorem later which assumes this fact.

We see that $\overline{H_t}$ now naturally breaks up into $\overline{H_t} := \sum (X_{i,t} + Y_{i,t})$, where $$X_{i,t} = -\frac{\partial^2}{\partial x_i^2} + t^2\lambda_i^2 x_i^2,~~~~~Y_{i,t} = t\lambda_i \cdot [(a^i)^*,a^i].$$ Notice that $X_{i,t}$ is just the standard expression for a harmonic operator, which we studied in the previous blog post (up to scaling by $|\lambda_i|$). Although $X_{i,t}$ acts on a function of $n$ coordinates, it only sees the $x_i$ coordinate. Hence, the operators $X_{i,t}$ all commute with each other, and it is easy to see that one may write any $0$-form eigenstate of the operator $X_t := \sum_i X_{i,t}$ as $$\psi(x_1,\ldots,x_n) = \psi_1(x_1)\cdots\psi_n(x_n),$$ where $\psi_i$ is an eigenstate for $X_{i,t}$. In particular, we find that $X_t$ is itself diagonalizable with basis given by letting $\psi_1,\ldots,\psi_n$ be the standard basis elements for the simple harmonic oscillator in dimension $1$. For $k$-forms, we simply use these as coefficients of the standard $k$-forms $dx_I$ with $|I| = k$.

Each $X_{i,t}$ and $Y_{j,t}$ commute with each other, and therefore, $X_t$ and $Y_{j,t}$ all commute. Since $X_t$ is diagonalizable, this amounts to the fact that each $Y_{j,t}$ preserves the eigenspaces of $X_t$. Since $Y_{j,t}$, or rather its restriction to each eigenspace of $X_t$, is skew-adjoint, it is diagonalizable on each eigenspace, and so $X_t$ and each $Y_{j,t}$ is simultaneously diagonalizable. But also the $Y_{j,t}$ themselves commute, so actually $X_t$, $Y_{1,t},\ldots, Y_{n,t}$ are all simultaneously diagonalizable.

Exercise: Verify that $[X_{i,t},Y_{j,t}] = 0$ and $[Y_{i,t},Y_{j,t}] = 0$. (We have already used that $[X_{i,t},X_{j,t}] = 0$ in our discussion of the eigenstates of $X_t$.)

The eigenvalues of $Y_{i,t}$ are $\pm t\lambda_i$, and so it follows from simultaneous diagonalizability that the eigenvalues of $\overline{H_t}$ are given by expressions of the form $$t\sum_{i=1}^{n}\left(|\lambda_i|(1+2n_i) + s_i\lambda_i\right),$$  where $s_i = \pm 1$. Notice that by checking the action of $Y_{i,t}$ on $k$-forms, one may see that exactly $k$ of the $s_i$ are equal to $+1$, while the rest are $-1$.

Exercise: Verify this fact about the number of positive $s_i$.

We see that for $t$ large, most of the eigenvalues are themselves going to be large, since the numbers $|\lambda_i|(1+2n_i) \pm \lambda_i$ are all nonnegative. The only way to not get a large eigenvalue is for all of these to be zero, which occurs when each $n_i = 0$ and when $s_i = - \mathrm{sgn}(\lambda_i)$. In particular, we see that the number of positive $s_i$ is the number of negative $\lambda_i$, and hence, our zero-energy eigenstates for $\overline{H_t}$ in a neighborhood of a critical point of index $k$ must be a $k$-form. Furthermore, this eigenstate is actually completely determined up to scaling, as the fact that $s_i$ is positive for negative $\lambda_i$ singles out the $k$-form $dx^I$ where $I$ consists of those indices with negative $\lambda_i$. Stated precisely, since $H_t \approx \overline{H_t}$, we arrive at the following result.

Theorem (Weak Morse Inequalities): The dimension of the $k$th de Rham cohomology $H^k_{\mathrm{dR}}(M)$ is bounded above by the number of critical points of index $k$ of any Morse function $f$ on $M$.

Proof: The $0$th de Rham cohomology is naturally associated with the zero energy eigenspace of $H_t$, as discussed. Given the approximation $H_t \approx \overline{H_t}$ near the critical points, it follows that every zero-energy eigenstate for $H_t$ which is a $k$-form is approximately a low-energy eigenstate for $\overline{H_t}$ near each critical point of the Morse function $f$ which, again, is a $k$-form. But there is a unique such eigenstate near each critical point of index $k$ (and around no other critical points).

Definition: The dimension of the $k$th de Rham cohomology is often called the $k$th Betti number $\beta_k(M)$.

Remark: One point in the above proof is that every zero-energy state of $H_t$ must be a zero-energy state of $\overline{H_t}$, since all low energy states of $\overline{H_t}$ are automatically zero-energy states. The converse is not true! In particular, a zero-energy state of $\overline{H_t}$ will correspond, in general, to a low-but-not-necessarily-zero-energy state of $H_t$. Otherwise, the (weak) Morse inequality would be an equality for any Morse function, but this is certainly impossible, since at the very least we can create and cancel critical points, and so the number of critical points of a Morse function is not an invariant of a manifold. (One has, however, that the minimum number of critical points of a Morse function is an invariant, and it is not necessarily equal to the sum of the Betti numbers!)

The weak Morse inequality was already known from standard Morse theory techniques, and is due to Morse. A beautiful exposition can be found in Part I Section 5 of Milnor's Morse Theory. In fact, one may take the proof of the so-called strong Morse inequality as partial motivation for what comes next.

Theorem (Strong Morse Inequalities): Let $C_k(f)$ be the number of index $k$ critical points of a Morse function $f$ on a manifold $M$. Then for each $\ell \geq 0$, $$\beta_{\ell}(M) - \beta_{\ell-1}(M) + \cdots + (-1)^{\ell}\beta_0(M) \leq C_{\ell}(f) - C_{\ell-1}(f) + \cdots + (-1)^{\ell}C_0(f).$$

Exercise: Prove that in order to prove the strong Morse inequalities, it suffices to prove that there exists a cochain complex $$0 \rightarrow X_0 \xrightarrow{\delta} X_1 \xrightarrow{\delta} \cdots \xrightarrow{\delta} X_n \rightarrow 0$$ where each $X_k$ is a vector space with $\delta^2 = 0$ such that

• $\mathrm{dim}(X_k) = C_k(f)$
• the cohomology of the cochain complex (i.e. $\mathrm{ker}(\delta)/\mathrm{im}(\delta)$) of the has rank given by the Betti numbers

This is not quite the proof that Milnor writes in the book Morse Theory, but it is the proper phrasing of the fundamental idea. We see immediately that the existence of the Morse-Smale-Witten complex, as described in the first blog post, implies the strong Morse inequalities.

Remark: Due to some lack of foresight, the Morse-Smale-Witten complex I described in the first post is actually dual to the one discussed here. In the first post, the differential decreased degree; here it increases degree. As everything can be dualized without much problem, I hope you will forgive this inconvenience.

There is now a natural choice for the various vector spaces $X_k$. For each critical point, we have a zero-energy eigenstate of the operator $\overline{H_t}$, which is a $k$-form for the critical points of index $k$. These correspond to some low-energy states of $H_t$ (but to reiterate, these states may have non-zero energy). We let these low-energy eigenstates for $H_t$ be a basis for $X_k$. Then we may take $\delta = d_t$. Indeed, $d_t$ commutes with $H_t$ and hence preserves the eigenspaces for $H_t$, and also $d_t^2 = 0$, so this forms a cochain complex as desired. It suffices to verify the two properties of the previous exercise. The first is obvious by construction. The second follows by similar finagling as in the Hodge theory section.

Exercise: Prove the second property, assuming the fact that the Hodge decomposition theorem holds with respect to $d_t$ and $d_t^*$. Hint: Prove that the eigenstates of non-zero energy are $L^2$-orthogonal to the twisted harmonic forms.

Given our natural basis for $X_k$ just described, one asks the natural question of what $d_t \colon X_k \rightarrow X_{k+1}$ looks like with respect to this basis. The basis is naturally indexed by the critical points of $f$ themselves, and we write $|p\rangle$ for the basis element corresponding to the critical point $p$. (This notation is referred to as a ket, and is surprisingly natural for reasons we will not see.) Then the data of $d_t$ is a collection of numbers $c(p,q)$ for each pair of critical points $p$ and $q$ with $\mathrm{ind}(p) - \mathrm{ind}(q) = 1$, i.e. defining $$d_t | q \rangle = \sum_{p} c(p,q)\cdot |p\rangle.$$

Here is where we would need to delve into the physics literature further, and in particular the path integral formulation of this supersymmetric quantum mechanical system, but it turns out that these values $c(p,q)$ should be seen as transmission coefficients for tunneling phenomena. In particular, they represent some sort of probability that the state $|q \rangle$ will tunnel into the state $|p \rangle$, completely analogously to how we described tunneling for free particles. If you believe this statement, then what should the probability be? Recall that from the WKB approximation, in the $1$-dimensional setting, we found a transmission probability of the form $e^{-2\gamma}$, where $$\gamma \approx \int \sqrt{2(V-E)},$$ where the integral was over a path from where our free particle started to where it was transmitted. In our setting, we have low energy states, so $E \approx 0$. Additionally, we are using a different normalization (the Laplacian does not have a factor of $\frac{1}{2}$ in front of it), so in the end, we should expect our transmission probability is something like $$\gamma \approx \int_{\mathcal{P}(q,p)} \sqrt{V}$$ where we integrate over paths from $q$ to $p$, which we've denoted by $\mathcal{P}(q,p)$. This space of paths is infinite-dimensional, a typical problem in the land of path-integrals, but there are standard ways to ignore this problem, and they lead to the fact that $\gamma$ can be computed by simply summing over the leading contributing terms, by which we mean the paths of minimal energy, which in this work out to be paths given by gradient flow! In particular, we should take $$\gamma = \sum_{[\phi] \in \mathcal{M}(p,q)} \int_{-\infty}^{\infty} \sqrt{V(\phi(x))}~d\phi(x),$$ where $\mathcal{M}(p,q)$ is the moduli space of negative gradient trajectories. Meanwhile, notice that $$V(x) \approx t^2\|df\|^2,$$ where we assume $t$ is large so that the other term with a single factor of $t$ is very small. Intuitively, this means that for a gradient flow trajectory $\phi$, we have $$\int_{-\infty}^{\infty} \sqrt{V(\phi(x))}~d\phi(x) = \int_{-\infty}^{\infty}t \frac{df(\phi(x))}{dx}~dx = t(f(p)-f(q))$$ (where the equality should be taken loosely). We arrive at the result $$c(p,q) = n(p,q) \cdot e^{-2t(f(p)-f(q))},$$ where $n(p,q)$ is the number of negative gradient flow trajectories from $p$ to $q$ (counted with sign).

Up to these last hand-wavy steps, we really have recovered the Morse-Smale-Witten complex as desired!

Remark: The last step might seem to be the most difficult, but that is more due to the exposition than due to the fact of the matter. My understanding is that the tunneling computation, which is really an instanton computation from the proper path-integral perspective, was already standard in physics at the time of Witten's paper. My hope is that the intuition I have tried to derive from the comparison to the very simple tunneling computation coming from the $1$-dimensional WKB approximation described in the previous post is sufficient for some level of belief that the ideas leading to the Morse-Smale-Witten complex are quite natural from the physical perspective.