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math214:final-solution

## Final Solution

$$\gdef\gfrak{\mathfrak g}$$

First, I have to admit that I stealed some problems from Prof Valentino Tosatti's final problems, with some modifications (problem 1,2,3). Here is his course website and the final with solution.

I will write not below the most concise solution, but some remark and explanations along the way.

### Problem 1

1. (15 pt) Let $G$ be a Lie group, $\gfrak = T_e G$ its Lie algebra. Let $TG$ be identified with $G \times \gfrak$ by $$G \times \gfrak \to TG, \quad (g, X) \mapsto (L_g)_* X$$ Endow $TG$ with the natural induced Lie group structure, $$\rho: TG \times TG \to TG$$ such that if $\gamma_1, \gamma_2: (-\epsilon, \epsilon) \to G$ are two curves in $G$, then $$\rho(\dot \gamma_1(0), \dot \gamma_2(0)) = (d/dt)|_{t=0} (\gamma_1(t) \gamma_2(t)).$$ Write down the product law of $TG$ using identification with $G \times \gfrak$, i.e. $$(g, X) \cdot (h, Y) = ?$$

Solution: First consider the product of curves $$g e^{t X} h e^{t Y} = gh \cdot h^{-1} e^{t X} h e^{t Y},$$ this is a curve passing through point $gh \in G$ at $t=0$. We use left-multiplication by $(gh)^{-1}$ to pull-back this curve so that it passes through $e$ at $t=0$, then take its derivative $$\frac{d}{dt}|_{t=0} (h^{-1} e^{t X} h e^{t Y}) = Ad_{h^{-1}} X + Y$$ where we used Lebniz rule. If you wish, you can think of this as matrix multiplication, i.e., pretend $X, Y$ as $n$ by $n$ matrices and $h$ as an invertible matrix.

### Problem 2

2. (15 pt) Let $\C$ acts on $\C^p \RM \{0\} \times \C^q \RM \{0\}$ by $$t \cdot (z_1, \cdots, z_p; w_1, \cdots, w_q) \mapsto (e^{it} z_1, \cdots, e^{it} z_p; e^t w_1, \cdots, e^t w_q)$$ Show that the action is free, and the quotient is diffeomorphic to $S^{2p-1} \times S^{2q-1}$.

Solution: See Tosatti's solution above.

### Problem 3

3. (20 pt) Let $M$ be a smooth manifold, $\nabla$ be a connection on $TM$. Recall the torsion is defined as $$T: TM \times TM \to TM, \quad T(X, Y) = \nabla_X Y - \nabla_Y X - [X, Y].$$

• (10 pt) Show that the following formula defines a new connection $$\widetilde \nabla_X Y = \nabla_X Y - (1/2) T(X, Y)$$ where $X, Y$ are any vector fields.
• (5 pt) Show that the new connection is torsionless.
• (5 pt) Let $G$ be a Lie group. Let $\nabla$ be the flat connection on $TG$ where the left-invariant vector fields are flat sections. Compute the torsion of this connection.

Solution: For part 1 and 2, see Tosatti's solution above. For part 3, we first need to recall that the torsion tensor $T(- , -)$ is $C^\infty(M)$ linear in both slots, i.e., for whatever vector fields $X, Y$, and smooth functions $f,g$, we always have $$T(fX, gY) = fg T(X, Y)$$ in other words, $T(X, Y)|_p$ only depends on the value of the vector fields $X|_p$ and $Y|_p$ pointwise at $p$, not depends on how $X$ and $Y$ varies near $p$. This is what it means to be a tensor (see Lee or Nicolascu for the discussion).

Hence, if someone write $T(X, Y) = -[X, Y]$ for any vector field $X$ and $Y$, then it is automatically wrong, because $-[fX, gY] \neq -fg [X, Y]$.

For a point $p \in G$, to specify that $T_p$ is, we only need to ask how $T_p$ acts on two basis vectors in $T_p G$. Let $E_1, \cdots, E_n$ be a set of basis vectors in $T_e G$, we extend them to be a left-invariant vector fields $X_1, \cdots, X_n$ on $G$, then $\{X_i|_p\}$ is a basis on $T_p G$. We have then $$T_p (X_i|_p, X_j|_p) = T(X_i, X_j)|_p = (\nabla_{X_i}(X_j) - \nabla_{X_j}(X_i) - [X_i, X_j] )|_p = - [X_i, X_j]|_p.$$ Here, these $\nabla_{X_i}(X_j)$ term vanishes because $X_j$ is a flat section with respect to the connection $\nabla$.

### Problem 4

4.(15 pt) Let $\pi: S^3 \to S^2$ the Hopf fibration. Let $\omega$ be a 2-form on $S^2$ such that $[\omega] \in H^2(S^2)$ is non-zero.

• (10 pt) Show that there exists a 1-form $\alpha \in \Omega^1(S^3)$, such that $$d\alpha = \pi^* \omega.$$
• (5 pt) Suppose $\omega$ is the volume form on $S^2$ from the round metric, can you give an explicit construction of such an $\alpha$ on $S^3$, and $\alpha \wedge d\alpha$ is a non-vanishing 3-form on $S^3$?

Solution: part (1) should be easy, since $\pi^*(\omega)$ is a closed 2-form in $\Omega^2(S^3)$. Since $H^2(S^3)=0$, hence closed 2-form are also exact, that means, there exist some $\alpha$ such that $d \alpha = \pi^* \omega$.

part (2) is kind of evil, admittedly. If you sit down and take a piece of paper and take coordinates on $S^2$ and $S^3$, you may still not be able to find the answer. The answer is $$\alpha = (x_1 dx_2 - x_2 dx_1 + x_3 dx_4 - x_4 d x_3)|_{S^3}$$ where $x_i$ are coordinate on $\R^4$, and we restrict this one-form to $S^3$.

The reason that I put this problem here is to introduce a few concepts: (… is final exam a good place to introduce new concepts? anyway..)

• contact structure and symplectic structure. $\omega$ is a symplectic two-form on $S^2$, and $\alpha$ is a contact 1-form on $S^3$. see wiki page and here
• There is a general story of Kahler quantization or geometric quantization behind the scene. Read about it from John Baez.

And one remark: the word 'non-vanishing' means the 3-form $\alpha \wedge d\alpha$ is non-zero at each point $p \in S^3$.

### Problem 5

5. (10 pt) Let $(M, g)$ be a Riemannian manifold, a closed geodesic is a geodesic $\gamma: [0, 1] \to M$ such that $\gamma(0)=\gamma(1)$ and $\dot \gamma(0) = \dot \gamma(1)$.

• (7 pt) Let $M$ be a genus $g\geq 1$ surface, smooth, compact without boundary, orientable 2-dimensional manifold and $g$ any smooth Riemannian metric. Is it true that there is always exists a closed geodesic on $M$?
• (3 pt) Let $M = S^2$ and $g$ any smooth Riemannian metric. Is it true that there is always exists a closed geodesic on $M$? Try your best to give an argument.

Solution: (a) If a manifold $M$ has a non-trivial $\pi_1(M)$, ie., there exists a loop on $M$ that cannot shrink to a point, then we can take that loop, shrink it as much as possible, then we get a geodesic. The condition on $g \geq 1$ is to ensure that $\pi_1(M)$ is non-trivial.

(b) Of course, you can use google to get '3 geodesic theorem' and the Lyusternik–Fet theorem. But it is more fun to come up with some sketch of idea yourself. Recall, if the metric on $S^2$ is the round metric, then great circles are geodesic, however it is unstable, and a slight perturbation will make it either slip off to the left or to the right. But in general, how to look for the analog of great circle?

The idea is to consider an interval worth of loops, parametrized by $s \in [0,1]$, for example, take various horizontal slices of the $S^2=\{x^2+y^2+z^2=1\}$, where $s=0$ corresponds to the constant loop at south-pole, $s=1$ the constant loop at north pole, and $s=1/2$ the equator. Of course, for generic metric, the equator will not be a geodesic. Then, we try to 'pull the rope taut' simultaneously for all the loops, or run the 'curve-shortening-flow' on the free loop space $L(S^2):=Map(S^1, S^2)$. The point is that, there will be some loop that 'hangs' in the middle, and do not shrink to a point.

OK, now what about $S^3$? To learn this subject properly, one can read this book Lectures on Closed Geodesics.

### Problem 6

6. (15 pt) Let $K \In \R^3$ be a knot, that is, a smooth embedded submanifold in $\R^3$ diffeomorphic to $S^1$.

• (10 pt) Can you construct a geodesically complete metric on $M = \R^3 \RM K$? i.e. a metric such that for any $p \in M$, $v \in T_p M$, the geodesics with initial condition $(p,v)$ exists for inifinite long time? Describe your construction explicitly.
• (5 pt) Assume such metric exists, prove that for any point $p \in M$, there are infinitely many distinct geodesics $\gamma: [0,1] \to M$ with $\gamma(0)=\gamma(1)=p$.

Solution: One can make the metric blow-up near the knot $K$, so that the distance to the knot will be $\infty$. How to make that happen? Consider how to put a metric on the open interval $(0, 1)$ to make the metric complete? We can write $$g = f(x) dx^2$$ for some smooth function $f(x) > 0$ on $(0,1)$, and we want to make sure $$\int_0^{1/2} \sqrt{f(x)} dx = \infty, \quad \int_{1/2}^1 \sqrt{f(x)} dx = \infty$$ Then, if you recall that $\int_0^\epsilon (1/x) dx = \infty$, then you can cook up some $f(x)$ so that $f(x) \sim 1/x^2$ near $x=0$ and $f(x) \sim 1/(1-x)^2$ near $x=1$. (Of course, to feel safe, you can let $f(x) \sim 1/|x|^p$ for $p > 2$ near $x=0$).

part (2) again is about $\pi_1(\R^3 \RM K)$ being non-trivial. Similar to problem 5 part (a).

### Problem 7

7. (10 pt) Let $G = SU(2)$. Let $\nabla^{L}$ (resp. $\nabla^{R}$) be the flat connection on $TG$, where the flat sections are left (resp. right)-invariant vector fields. Prove that there is no 1-parameter family of flat connections $\nabla^{(t)}$ connecting $\nabla^{L}$ and $\nabla^{R}$, i.e. $\nabla^{(0)} = \nabla^{L}$ and $\nabla^{(1)} = \nabla^{R}$

Solution: Let's consider a general Lie group $G$ first. Given an element $E \in T_e G =: \gfrak$, then we can locally extend $E$ to flat section $X$ in a open neighborhood of $e$, using parallel transport. If furthermore $G$ is simply connected, then we can get global flat sections $X$ extending $E$. By replacing $G$ with its universal cover, we may assume $G$ is simply connected. Then, we have

Lemma: If $G$ is simply connected, then we have a bijection of sets $$\{ \z{Flat connections on TG} \} \leftrightarrow \{ \z{ trivializations \Phi: TG \to G \times \gfrak  , such that \Phi on T_e G is the identity map. \}$$ Both space has natural topology (for example, the space of flat section induce topology from the space of all connections) and this above bijection is actually a homeomorphism.

Then, finding a path connecting left and right flat connection $\nabla^L$ and $\nabla^R$ is equivalent to finding a path $\Phi_t$ between left and right trivializations $\Phi^L$ and $\Phi^R$ of $TG$. Consider the 'difference' of two trivializations: $$\Phi_t \circ \Phi_0^{-1}: G \times \gfrak \to G \times \gfrak$$ that fixes the $G$ slot, and is linear on the $\gfrak$ slot. Hence we get a family of maps $$\varphi_t: G \to GL(\gfrak)$$ such that $\varphi_0$ is the constant map $G$ to $Id \in GL(\gfrak)$, and $\varphi_1: G \to GL(\gfrak)$ is the adjoint representation $Ad: G \to GL(\gfrak)$. Then the question boils down to: is the adjoint representation homotopic to the constant map?

We claim that, in the case $G = SU(2)$ (or more generally compact semi-simple Lie group), there is a non-trivial $\omega_3 \in \Omega^3( GL(\gfrak))$, such that $\varphi_1^*(\omega_3)$ is the Cartan 3-form (up to a constant) (see Nicolascu page 280, Prop 7.4.21, or Homework 12). Since the cohomology class of $[\varphi_t^*(\omega_3)]$ should be invariant, hence $[\varphi_t^*(\omega_3) ] \neq 0$ in $H^3(SU(2))$, which contradict with $\varphi_0$ being a constant map. Thus, the map $Ad: G \to GL(\gfrak)$ is not homotopic to a constant map.

The form $\omega_3$ on $GL(\R^N)$ can be defined as following $$\omega_3 := tr( g^{-1} dg \wedge g^{-1} dg \wedge g^{-1} dg).$$

Some of you tried to construct a linear interpolation of connections $\nabla_t = t \nabla^L + (1-t) \nabla^R$, and find that it is not flat for $t \in (0,1)$. That is only circusmstancial evidence that there are no path within flat connections between $\nabla^L$ and $\nabla^R$, but not a proof.

### Afterwords

This final has some challenging problems, e.g. problem 4-b, 5-b, 7. Hence it is OK if you find it hard, or cannot do all the problems. I hope this final is thought-provoking enough. And I also hope this class serves as a beginning as a journey to the wonderful subject. Eventually, what matters is the discussion, the examples, the intuitions and the guesses (and of course theorems and proofs).

math214/final-solution.txt · Last modified: 2020/05/17 16:23 by pzhou