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.


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.

  • $\begingroup$ Can you say what $df(M)$ is meant to denote? $\endgroup$ – John Hughes Nov 28 '18 at 14:02
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ 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

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 $\mathbb{R}^2$.


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