10-29 Notes

\section{Morse-Bott Functions, Continued}

We complete the discussion of Morse-Bott functions from the last lecture. Recall that we have a smooth finite dimensional manifold $M$, a Morse-Bott function $f:M \to \mathbb{R}$, $C$ the critical set for $f$, and $\gamma (t)$ a flow line for $\nabla f$. We know that in general $\lim_{t\to\infty} \textrm{dist}(\gamma (t), C)=0$. We would like to use this fact to prove the stronger result used in the previous lecture, namely, that in fact there exists $c\in C$ with $\lim_{t\to\infty}\gamma (t) = c$, and furthermore that the convergence is exponential. We begin by noting that the Morse-Bott condition ensures that there exists an open neighbourhood $U$ of $C$ such that for all $x\in U$, $\textrm{dist}(x,C)\leq |\nabla_x f|$. Every flow line is contained in $U$ for $t$ sufficiently large.

\begin{thm} Suppose that $\gamma (t) \subset U$ for all $t$, and that $f\equiv 0$ on the connected component of $C$ which $\gamma (t)$ approaches. Then there exists $C, \alpha >0$ such that

$$ f(\gamma (t)) \leq C e^{-\alpha t}$$

\end{thm}

\begin{proof}

Since $\gamma(t)$ is a flow line for $f$, we have $\frac{d}{dt}f(\gamma(t))=-|\nabla_{\gamma(t)}f|^2$. For $x\in U$ we have for some $\alpha > 0$

$$\alpha|f(x)|^2 \leq \textrm{dist}(x,C)^2 \leq |\nabla_xf|^2$$

and so

$$ \frac{d}{dt}f(\gamma(t) \leq -\alpha f(\gamma(t))$$

$$ \frac{1}{f}\frac{d}{dt}f(\gamma(t))\leq -\alpha$$

$$\ln(f(\gamma(t)-\ln(f(\gamma(0) \leq -\alpha t$$

$$f(\gamma(t)) \leq f(\gamma(0)) e^{-\alpha t}$$

so setting $C=f(\gamma(0))$ completes the proof

\end{proof}

Thus, we have

$$|\gamma(t+1)-\gamma(t)| \leq \int_t^{t+1} |\nabla_\gamma(s)f|ds \left(\leq \int_t^{t+1} |\nabla_\gamma(s)f|^2ds \right)^{\frac{1}{2}}$$

Recalling that $\frac{d}{dt}f(\gamma(t))=-|\nabla_{\gamma(t)}f|^2$ and the previous theorem, this in turn is less than or equal to $(Ce^{-\alpha t})^{\frac{1}{2}}$. Note that the same argument works for $t+\nu$ for any $0 < \nu < 1$. Iterating this estimate, we see that $|\gamma(t+n)-\gamma(t)| \leq C' e^{\frac{-\alpha t}{2}}$. Thus, $\gamma(t)$ is exponentially Cauchy and hence converges exponentially fast to some $c\in C$, as desired.

For the previous lecture, we also wanted to know that $\int_0^\infty |\dot{\gamma}|^2 e^{2\delta t}dt < \infty$ for $\delta$ sufficiently small. But this follows immediately by integration by parts from what we have already established, since this integral is equal to

$$\int_0^{\infty} -\dot{f}e^{2\delta t} dt = f(0) + \int_0^\infty \frac{f e^{2\delta t}}{\delta} dt $$

and the right hand side is finite for $\delta$ sufficiently small by the exponential convergence of $\gamma(t)$.

\section{Morse Stratification and Injectivity of the Euler Class}

We are now going to apply the foregoing theory to the following situation. Suppose that $X$ is a (not necessarily finite-dimenionsal) manifold, and $f$ a Morse-Bott function on $X$. We assume that $f$ induces a Morse stratification $X=\bigcup _i X_i$ on $X$, as in the finite dimensional case; the verification of this assumption for the Yang-Mills functional will occupy us for number of lectures. We assume there is also a group $G$ acting smoothly on $X$ preserving $f$. Then $f$ induces a Morse-Bott function on the Borel quotient $X_G := X \times EG /G$, and it makes sense to discuss the equivariant cohomology of the stratification. In other words, we get an exact sequence of the form

$$H^{*-1}_G(X_{j-1})\overset{\delta}{\to} H^{*}_G(X_j, X_{j-1}) \to H^{*}_G(X_j)$$

We would like to know when the second map is injective, since this will tell us how the cohomology of $X$ changes as we pass to higher strata. From the definition of the Morse statrification, for some critical subset $C$ of appropriate index the above sequence maps into

$$H^{*-1}_G (W^u_C-C)\overset{\delta}{\to} H^{*}_G(W^u_C, W^u_C - C) \to H^{*}_G(W^u_C)$$

such that the map $H^{*}_G(X_j, X_{j-1}) \to H^{*}_G(W^u_C, W^u_C - C)$ is an isomorphism. Thus, it would suffice to show that the second map here is injective to show that $H^{*}_G(X_j, X_{j-1}) \to H^{*}_G(X_j)$ is also injective. From the previous lecture we know that $W^u_C$ is the total space of $\nu_-$, the subbundle of the normal bundle to $C$ on which the quadratic form associated to $f$ is negative definite. By the Thom Isomorphism theorem, the last map is equivalent to

$$H^{*-n_-}_G(C)\overset{\cup e(\nu_-)}{\to} H^{*}_G(C)$$

where $n_-=\textrm{dim}(\nu_-)=\textrm{Index}(C)$ and $e(\nu_-)$ is the Euler class of $\nu_-$. In the finite dimensional case, cupping with the Euler class of a vector bundle can of course never be an injective map. However, we will show that in the case of the Yang-Mills action this map is indeed injective.

We now specialize to the Yang-Mills action over Riemann surfaces. Consider a Riemann surface $\Sigma$ of genus $g$, and set $X=\mathcal{A}=\mathcal{A}(2,1)$ the space of connections on the unique $U_2$ vector bundle $E$ with $< c_1(E), [\Sigma]> =1$, $f$ the Yang-Mills functional $\int_\Sigma |F_A|^2 dvol$, and $G=\mathcal{G}$ the gauge group of $E$. In fact, that is not quite right; instead we consider a fixed connection $\delta$ on the line bundle $\Delta = \det E$, and consider $\mathcal{A}^\delta,$ the set of connections with determinant $\delta$, and $\mathcal {G}^\delta$ the gauge transformations with trivial determinant.

We already know what the critical points of the Yang-Mills functional are; see the lecture notes for 10/15 for a picture. There are two classes of Yang-Mills connections: the minimal energy connections, which consist of irreducible connections (the quotient of which by $\mathcal{G}^\delta$ we have been calling $N^\delta(2,1$)); and the higher energy Yang-Mills connections, for which there exits a splitting $E=E_1\oplus E_2$ compatible with the connection. We define $C_i :=$ {Yang-Mills connections $A\in\mathcal{A}^\delta(2,1)$ compatible with a splitting $E=E_1\oplus E_2$ such that $<c_1(E_1), [\Sigma]> = i$}. Note that in such case we must have $<c_1(E_2), [\Sigma]> = 1-i$. The main theorem of this section is the following:

\begin{thm} On $C_i$, cupping with the Euler class $H^{*-n_-}_\mathcal{G^\delta}(C_i)\overset{\cup e(\nu_-)}{\to} H^{*}_\mathcal{G^\delta}(C_i)$ is injective on equivariant cohomology. \end{thm}

\begin{proof}

We need to look at $(C_i\times E\mathcal{G}^\delta )/\mathcal{G}^\delta$. There are at least two obstacles to doing so: $\mathcal{G}^\delta$ is a rather large group, its action on $C_i$ is not free but has a $U_1$ stabilizer. We deal with the first problem in the following manner: let $x_0\in \Sigma$ be any point, and define

$$ \mathcal{G}_0 = \{u\in \mathcal{G}^\delta\textrm{ } | \textrm { } u|_{x_0}=Id\}$$

We have a short exact sequence $1\to \mathcal{G}_0\to\mathcal{G}^\delta\to SU_2\to 1$. $\mathcal{G}_0$, however, acts freely on $\mathcal{A}^\delta$. Thus, if we set $\mathcal{B}^0 := \mathcal{A}^\delta/ \mathcal{G}_0$, we have $H^*_{\mathcal{G}^\delta} (\mathcal{A}^\delta) = H^*_{SU_2}(\mathcal{B}^0)$. In other words, we have reduced to the case of a finite dimensional group action.

We have the bundle $\pi: \nu_- \to C_i$, and for $A\in C_i$, $\pi^{-1}(A)=\textrm{Ker}(\overline{\partial}_A:\Omega^{1,0}(\Sigma, E_1^*\otimes E_2)\to \Omega^{1,1}(\Sigma, E_1^*\otimes E_2))$. Since $\mathcal{G}_0$ acts freely on $C_i$ and $\nu_-$, defining $\nu_-^0:=\nu_-/\mathcal{G}_0$ and $C_i^0 := C_i/\mathcal{G}_0$, we have an honest vector bundle $\nu_-^0 \to C_i^0$, with $SU_2$ acting on every term. We need to show that $e_{SU_2}(\nu_-^0)\in H^*_{SU_2}(C_i^0)$ is not a zero divisor.

To do so, note that there is a $U_1\subset SU_2$ which acts trivially on $C_i^0$. Thus, we have a map

$$C_i^0 \times BU_1 = (C_i^0\times ESU_2)/U_1 \overset{\rho}{\to} (C_i^0\times ESU_2)/SU_2 = (C_i)_{\mathcal{G}^\delta}$$

Thus, at the level of cohomology we get a map $\rho^*: H^*_{SU_2}\to H^*(C_i^0\times BU_1)$. One can show (and perhaps we will next time) that the image of this map is a summand in $H^*(C_i^0\times BU_1)$. Thus if we show that $\rho^*(e_{SU_2}(\nu_-^0))=e(\rho^*(\nu_-^0))\in H^*(C_i^0 \otimes BU_1)$ is not a zero divisor we will be done. We will show that in fact this class is not a zero divisor when restricted to the $BU_1$ factor.

For any connection $A\in C_i$, we have $e(\rho^*(\nu_-^0))|_{[A]\times BU_1} = (EU_1\times \mathbb{C}^N)/U_1$. Note that this is an honest vector bundle on $BU_1$. Recall that the $\mathbb{C}^N$ factor corresponds to $\textrm{Ker}(\overline{\partial}_A:\Omega^{1,0}(\Sigma, E_1^*\otimes E_2)\to \Omega^{1,1}(\Sigma, E_1^*\otimes E_2))$. For $u=e^{i\theta}\in U_1$, the action on the $\mathbb{C}^N$ factor is via $e^{-2i\theta}$, since this $U_1$ as the stabilizer of $A$ acts as

$$\left(\begin{matrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{matrix}\right)$$

on $E=E_1\oplus E_2$. We recall that $BU_1=\mathbb{C}P^\infty$, and so $H^*(BU_1)= \mathbb{Z}[u]$ where $|u|=2$, a polynomial algebra in one variable. From the fact that $U_1$ acts via $e^{-2i\theta}$ on the $\mathbb{C}^N$ factor, we deduce that $$e(\rho^*(\nu^0_-)|_{BU_1}) = 2^nu^n$$ which is obviously not a zero divisor. This completes the proof of the proposition.

\end{proof}

\section{Poincare Polynomial of $N^\delta(2,1)$}

We are now in a position to compute the rational cohomology of $N^\delta(2,1)$. In doing so, we invert the usual order of explanation in Morse theory: rather than using knowledge of the critical points to adduce information about the entire space, as is customary, we will instead use our knowledge of the algebraic topology of the entire space to deduce consequences for the topology of the critical points. We remind the reader that one significant assumption is still unproven: that the Yang-Mills functional induces a Morse-stratification on $\mathcal{A}^\delta$. We will spend the next several lectures paying the piper to justify this assumption.

For any space $X$, we recall the Poincare polynomial (or, more accurately, the Poincare power series)

$$P_t(X):=\Sigma_k t^k\textrm{dim}(H^k(X,\mathbb{Q}))$$

In our situation, we have a Morse stratification of $\mathcal{A}^\delta_{\mathcal{G}^\delta}=\mathcal{A}^\delta\times E\mathcal{G}^\delta/\mathcal{G}^\delta$. From the previous section, we know how the rational cohomology of $\mathcal{A}^\delta_{\mathcal{G}^\delta}$ changes as one passes from the $j-1$'st to the $j$'th strata; namely, since cupping with the Euler class on $C_i$ is injective the changes consists of simply taking the direct sum with the equivariant cohomology of $C_i$, shifted by the degree of the Euler class, which equals the index of the critical set $C_i$. In other words, we have

$$P_t(H_{\mathcal{G}^\delta}(\mathcal{A^\delta})) = \Sigma_{i=1}^\infty t^{n_-(i)}P_t(H_{\mathcal{G}^\delta}(C_i)) + P_t(N^\delta(2,1))$$

with the obvious notation $P_t(H_G(X))=P_t(X_G)$. There is one potential point of confusion in the previous equation: the space $N^\delta(2,1)$ appears on the right hand side, but $N^\delta(2,1) \subset \mathcal{A}^\delta/{\mathcal{G}^\delta}$, while we are working in the space $\mathcal{A}^\delta_{\mathcal{G}^\delta}$. Thus, the minimal energy set is not $N^\delta(2,1)$ but is the pre-image $\tilde{N} := \pi^{-1}(N^{\delta}(2,1))$ under the quotient map $\pi: \mathcal{A}^\delta_{\mathcal{G}^\delta} \to \mathcal{A}^\delta/{\mathcal{G}^\delta}$. Thus, what should really appear on the right hand side is $P_t(\tilde{N})$.

Fortunately, this is not an issue. Each connection $A \in N^\delta(2,1)$ has stabilizer $\{\pm Id\}=\mathbb{Z}_2$, and so by the discussion in lecture 8 the map $\pi: \tilde{N}\to N^\delta(2,1)$ is a fibration with homotopy fibre $B\mathbb{Z}_2 = \mathbb{R}P^\infty$. $\mathbb{R}P^\infty$, however, has the rational cohomology of a point, and so applying the Serre spectral sequence with coefficients in $\mathbb{Q}$ we see that $P_t(\tilde{N})=P_t(N^\delta(2,1))$, as desired.

We already know, from previous lectures, that $H^*_{\mathcal{G}^\delta}(\mathcal{A}^\delta, \mathbb{Q})$ is a (graded) free polynomial algebra, with one generator in dimension $2$, $2g$ generators in dimension $3$, and one generator in dimension $4$. Thus, we have

$$P_t(H_{\mathcal{G}^\delta}(\mathcal{A}^\delta))=\frac{1}{1-t^2}(1+t^3)^{2g}\frac{1}{1-t^4}$$

As for $H_{\mathcal{G}^\delta}(C_i)$, recall that $C_i/\mathcal{G}^\delta$ is homeomorphic to the Jacobian torus $H^1(\Sigma, \mathbb{R})/H^1(\Sigma,\mathbb{Z}) \tilde{=} T^{2g}$. Since elements of $C_i$ have stabilizer $U_1$, we have a fibration $\pi: (C_i)_{\mathcal{G}^\delta} \to T^{2g}$, with homotopy fibre $BU_1=\mathbb{C}P^\infty$. One can show that all of the differentials in the spectral sequence associated to this fibration are trivial, and so cohomologically $(C_i)_{\mathcal{G}^\delta}$ is a product of $\mathbb{C}P^\infty$ and $T^{2g}$. Thus we have

$$P_t(H_{\mathcal{G}^\delta}(C_i))= \frac{1}{1-t^2}(1+t)^{2g}$$

Finally, we computed in lecture 12 the Morse indices of the critical sets $C_i$; we showed that $n_-(i)=2(2i+g-1)$. Putting all of this together, we have

$$P_t(N^\delta(2,1)) = \frac{(1+t^3)^{2g}}{(1-t^2)(1-t^4)}-\Sigma_{i=1}^\infty t^{2(2i+g-2)}\frac{(1+t)^{2g}}{1-t^2}

= \frac{(1+t^3)^{2g}}{(1-t^2)(1-t^4)} - \frac{(1+t)^{2g}}{1-t^2}t^{2g}\frac{1}{1-t^4}$$

$$= \frac{(1+t^3)^{2g} - (1+t)^{2g}t^{2g}}{(1-t^2)(1-t^4)}$$

If $P_t(N^\delta(2,1))$ is the Poincare polynomial of a closed, finite-dimensional manifold, we would have

$$P_{1/t}(N^\delta(2,1)) = t^{-\dim(N^\delta(2,1))}P_t(N^\delta(2,1))$$

We compute

$$P_{1/t}(N^\delta(2,1)) = \frac{(1+t^{-3})^{2g} - (1+t^{-1})^{2g}t^{-2g}}{(1-t^{-2})(1-t^{-4})} = \frac{t^{-6g}((1+t^3)^{2g} - (1+t)^{2g}t^{2g})}{t^{-6}(1-t^2)(1-t^4)} = t^{-(6g-6)}P_t(N^\delta(2,1))$$

Thus, the Poincare polynomial for $N^\delta(2,1))$ is consistent with Poincare duality, and furthermore we see that the dimension of $N^\delta(2,1))$, if it is a manifold, is $6g-6$. We will show in the next section that it is in fact a manifold. Note that if $g=1$, i.e. $\Sigma$ is an elliptic curve, then the dimension is zero and there exists a unique minimal energy connection. We will exhibit this connection in the next lecture.