2
$\begingroup$

I have been asked to describe the tangents space $T_q(df(M))$ as a subspace of $T_q(T^*M)$ where $f\in C^\infty(M)$ and $df$ is a 1-form (or smooth section of $T^*M$).

Here, $df:M\rightarrow T^*M$ is a function and $df(M)$ is an image of $M$ under $df$. And this image is naturally a smooth manifold because $df$ is smooth embedding.

But I have no idea on it. What should I focus on to describe it? Even it is difficult to see why $df(M)$ is a smooth manifold. I will be very appreciate for any help. Thanks in advance.

Newly Added

I started to think how $T_q(T^*M)$ look like. Suppose that the dimension of $M$ is $n$.

Note that $\exists $ a chart $(U,\varphi=\{x^1,\dots, x^n\})$ such that $q=(p,v)\in T^*U$. Then we have $$v=\sum_{i=1}^{n}y^idx^i. $$

Thus, we can describe local coordinate function of $(p,v)$ as $(x^1,\dots,x^n,y^1,\dots, y^n)$. It implies that our basis of $T_p(T^*M)$ would be $$\left\{\frac{\partial}{\partial x^1},\dots ,\frac{\partial}{\partial x^n} ,\frac{\partial}{\partial y^1},\dots ,\frac{\partial}{\partial y^n} \right\} .$$

I think I can use this to describe $T_q(df(M))$. Currently my concern is that the dimension of $df(M)$ is $n$ since it is diffeomorphic to $M$. In other words, we should pick $n$ element that would be linear combination of the basis above. But I am still working on.

Added(12/5/2018)

Now, consider $T_qL=T_q(df(M))$. Let $q=(p,df(p))\in df(M)$. And note that $df(p)\in T^*M$ which means that \begin{align*} df(p)=\sum_{i=1}^n y^i dx^i \end{align*} where \begin{align*} y^i = df(p)\left( \frac{\partial}{\partial x^i } \right)=\frac{\partial f}{\partial x^i}\bigg|_{p}. \end{align*}

Observe that each $y^i$ is determined by $x^i$. In other words, each $y^i$ is a function of $x^i$. Therefore, $\forall p\in M$, we can represent $(p,df(p))$ as $(x^1,\dots, x^n)$. Thus, our basis for $T_qL$ would be \begin{align*} \left\{ \frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right\}. \end{align*}

Thus, $T_qL$ is a set of linear combination of the basis above.

But I am uncertain about it. I am trying to making sure if it is right.

$\endgroup$
  • $\begingroup$ Can you say what $df(M)$ is meant to denote? $\endgroup$ – John Hughes Nov 28 '18 at 14:02
  • 1
    $\begingroup$ @JohnHughes $df:M\rightarrow T^*M$ is defined by $$df(p)=(p,v)$$ where $v\in T_p^*M$. So, basically $df$ is a smooth function and $df(M)$ is the image of $M$ under $df$. $\endgroup$ – Lev Ban Nov 28 '18 at 15:05
  • $\begingroup$ No, i guess $df(p)=(p,df_p)\in T^*M$, where $df_p\in T^*_pM$ is the differential at $p$ $\endgroup$ – Federico Nov 28 '18 at 17:26
  • 1
    $\begingroup$ @PedroTamaroff I just meant $$df(p)=(p,v) $$ since $$T^*M=\coprod_{p\in M}T^*_pM$$ $\endgroup$ – Lev Ban Nov 28 '18 at 17:27
  • 1
    $\begingroup$ My advice is to consider the map $F=df:M\to T^*M$ and try to compute its differential $dF$ $\endgroup$ – Federico Nov 28 '18 at 17:29
0
$\begingroup$

Let $(U,\varphi=\{x^i\})$ be the local chart where $p\in U$ and $q=(p,v)\in T^*U$. Then $$ df = \sum_{i=1}^{n}\frac{\partial f}{\partial x^i} dx^i.$$

Then we have the local coordinate $\left\{ x^i,y^i=\frac{\partial f}{\partial x^i} \right\}$ of $T^*M$. Then

$$T_qL=span\left\{ \frac{\partial}{\partial x^i} + \sum_{k=1}^{n}\frac{\partial^2f}{\partial x^i\partial x^k} \frac{\partial}{\partial y^k} \right\}_{i=1}^{n}. $$

In order to understand, think about easy example when $f:\mathbb{R}\rightarrow \mathbb{R}$. If we look at the graph of $f$, for given $p\in \mathbb{R}$, the graph of tangent line at $p$, $$y=f'(p)(x-p)+f(p) $$ is $span\left\{ \overrightarrow{e_1}+\frac{\partial f}{\partial x}\overrightarrow{e_2} \right\}$ where $\{e_1,e_2\}$ is a standard basis of $\mathbb{R}^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.