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I am stuck on the following problem.

Let $\omega$ be the canonical symplectic structure. Let $\alpha$ be a differential 1-form on $M$, and let $$L=\{(p,\alpha(p)):p\in M\}\subseteq T^*M.$$ Consider a smooth map $i:M\rightarrow T^*M$ defined by $i(p)=(p,\alpha(p)).$

Show that the $i^*\omega=-d\alpha.$

I am totally stuck at this problem for several hours. The only thing I can think is that I should show that $$i^*\omega(p,q)=-d\alpha(p,q).$$ Actually, even I am not sure about this. I hope I could get some help from here. Any help would be appreciated. Thank you in advance.

Lev Bahn
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2 Answers2

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As I understand, the canonical symplectic structure is defined on $ T^*M $. So the pullback, $ i^* \omega $ is a $ 2 $-form on $ M $ and has to be evaluated at a point $ p \in M $ (as opposed to what you write as $ i^* \omega (p,q) $; if you meant $ (p,q) $ is a point of $ T^*M $, then that's not the case). The details are somewhat computational as follows:

For $ p \in M $, let $ U $ be a co-ordinate chart with co-ordinates $ x_1, x_2, \cdots, x_n $. We have the bases $ \left \{\frac{\partial}{\partial x_i} \right\}_{1 \le i \le n} $ for $ TU $ and $ \{ dx_i \}_{1 \le i \le n } $ for $ T^*U $. Then we have the co-ordinates $ x_1, \cdots , x_n, y_1, \cdots, y_n $ on the cotangent bundle around $ (p, \alpha(p)) $. Then the canonical symplectic form on $ T^*M $ is $ \omega = \sum_i (dx_i \wedge dy_i) $. Suppose $ \alpha = \sum_i a_i dx_i $ on $ U $. Then it is not hard to show that $$ di_p \left( \frac{\partial}{\partial x_j} \right) = \frac{\partial}{\partial x_j} + \sum_i \frac{\partial a_i}{\partial x_j} \cdot \frac{\partial}{\partial y_i} $$

Now, $$ (i^* \omega)_p \left(\frac{\partial}{\partial x_j}, \frac{\partial}{\partial x_k} \right) = \omega_{(p, \alpha(p))} \left( di_p \left( \frac{\partial}{\partial x_j} \right), di_p \left( \frac{\partial}{\partial x_k} \right) \right) = \sum_i (dx_i \wedge dy_i) \left( di_p \left( \frac{\partial}{\partial x_j} \right), di_p \left( \frac{\partial}{\partial x_k} \right) \right) $$ You have to check the indices $ j,k $ only and after a short computation (if I've got my signs correct) this is equal to $ -\frac{\partial a_k}{\partial x_j} + \frac{\partial a_j}{\partial x_k} $. After this it is easy to check that this indeed equals $ -(d \alpha)_p \left(\frac{\partial}{\partial x_j}, \frac{\partial}{\partial x_k} \right) $ which verifies the assertion.

hellHound
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  • Thank you for the answer. Could you give me more detail on $$di_p\left( \frac{\partial }{\partial x_i}\right) =\frac{\partial}{\partial x_j}+\sum_{i}\frac{\partial a_i}{\partial x_j}\frac{\partial}{\partial y_i}?$$ I cannot see how the coefficients $a_j$ of $\alpha$ went there. – Lev Bahn Dec 04 '18 at 23:13
  • Oh! never mind! I figured out! Thank you! :) – Lev Bahn Dec 05 '18 at 17:48
  • Yes, it is just the chain rule. Glad my answer helped. – hellHound Dec 05 '18 at 18:41
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I got a different idea about this problem!

Let $\pi : T^*M\rightarrow M$ defined by $(p,v)\mapsto p$ and let $\tau_{(p,v)}=\pi^*_{(p,v)}(v)$.

And note that $$\pi_{(p,v)*} : T_{(p,v)}(T^*M)\rightarrow T_pM \\ \pi^*_{(p,v)}: T^*_pM\rightarrow T^*_{(p,v)}(T^*M). $$

Then we can define $\tau(p,v):= \pi^*_{(p,v)}(v) $ where $\omega = -d\tau$. And further note that $\alpha \in \Omega^1(M)$, so $$\alpha : M\rightarrow T^*M. $$

Now, observe that, since $\pi\circ i =Id_M$, we have $i^*\circ \pi^* = Id^*_M$. Observe that $$(i^*\tau(p))=i^*(\tau(p,\alpha(p)))=i^*\circ\pi^*_{(p,\alpha(p))}(\alpha(p))=\alpha(p)\hspace{2cm} \forall p\in M.$$ Thus, $i^*\tau=\alpha.$ Lastly, observe that $$ i^*\omega = -i^*d\tau=-d(i^*\tau)=-d(\alpha)=-d\alpha. $$

Then we are done.

Lev Bahn
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