# Tangent space $T_q(df(M))$ as a subspace of $T_q(T^*M)$

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.

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.

• Can you say what $df(M)$ is meant to denote? – John Hughes Nov 28 '18 at 14:02
• @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 $df$. – Lev Ban Nov 28 '18 at 15:05
• No, i guess $df(p)=(p,df_p)\in T^*M$, where $df_p\in T^*_pM$ is the differential at $p$ – Federico Nov 28 '18 at 17:26
• @PedroTamaroff I just meant $$df(p)=(p,v)$$ since $$T^*M=\coprod_{p\in M}T^*_pM$$ – Lev Ban Nov 28 '18 at 17:27
• My advice is to consider the map $F=df:M\to T^*M$ and try to compute its differential $dF$ – 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 $$\mathbb{R}^2$$.