# Differential map between tangent spaces

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1. May 28, 2015

### "Don't panic!"

I've been struggling since starting to study differential geometry to justify the definition of a one-form as a differential of a function and how this is equal to a tangent vector acting on this function, i.e. given $f:M\rightarrow\mathbb{R}$ we can define the differential map $$df(\mathbf{v})=\mathbf{v}(f)$$ where $\mathbf{v}\in T_{p}M$ (suppressing the $p"$ notation in the above definition for brevity).
Does this follow from a more general case? That is, if one has two manifolds $M,N$ then, given a smooth function, $f:M\rightarrow N$ that maps between the two, one can define a differential map $$df_{p}:T_{p}M\rightarrow T_{f(p)}N$$ that maps vectors in the tangent space $T_{p}M$ to a point $p\in M$ to tangent vectors in the tangent space $T_{f(p)}N$ to the point $f(p)\in N$. Given this we can define such a map by the way a "new" vector $df_{p}(\mathbf{v})\in T_{f(p)}N$ (where $\mathbf{v}\in T_{p}M$) acts on a function $g:N\rightarrow\mathbb{R}$ defined on $N$, i.e. $$\left(df_{p}(\mathbf{v})\right)(g)=\mathbf{v}_{p}(g\circ f)$$ In other words, the action of $df_{p}(\mathbf{v})$ on $g$ in $N$ should be equal to the action of $\mathbf{v}$ on $g\circ f$ in $M$.
Given this, then the first definition I gave is just a special case, in which $N=\mathbb{R}$ and $df_{p}:T_{p}M\rightarrow T_{f(p)}\mathbb{R}\cong\mathbb{R}$ becomes a one-form in the dual space $T^{\ast}_{p}M$ to the tangent space $T_{p}M$ at $p\in M$. Also, we find that $g:\mathbb{R}\rightarrow\mathbb{R}$ and so $g=\text{id}$ is simply the identity map. Hence we find, $$\left(df_{p}(\mathbf{v})\right)(\text{id})\equiv df_{p}(\mathbf{v})=\mathbf{v}_{p}(\text{id}\circ f)=\mathbf{v}_{p}(f)$$
Would this be correct, or am I missing something?

2. May 28, 2015

### Ben Niehoff

I think that's correct. The map you are calling $d f_p$ is also called the pushforward $f_*$.

3. May 28, 2015

### "Don't panic!"

Is that in the general case I gave? As in the particular case $N=\mathbb{R}$, isn't $df$ just a one-form?
Also, how does the exterior derivative of $f$ relate to this definition, is it just that the two describe the same object in the specific case where $N=\mathbb{R}$ (i.e. $\text{d}f\vert_{p}=df_{p}$ when $N=\mathbb{R}$)?

Last edited: May 28, 2015
4. May 28, 2015

### mathwonk

not sure what you are asking but the exterior derivative of a function is just the differential, i.e. it is the one form whose value at p is df(p).

5. May 29, 2015

### "Don't panic!"

Sorry I didn't word it very well, what I was meaning was does the differential map $df_{p}:T_{p}M\rightarrow T_{f(p)}N$ describe the differential of a function in the specific case where $N=\mathbb{R}$, i.e. does the differential map coincide with the exterior derivative in this case?

Last edited: May 29, 2015
6. May 29, 2015

### Ben Niehoff

Yes, the exterior derivative of a function is just the differential of the function. That's why they both use the same notation, $df$.

7. May 29, 2015

### lavinia

Yes. The differential is defined in the same way for a differentiable function between manifolds. This is a 1 form with values in a vector bundle. In Euclidean space it is a vector valued 1-form. You can not generalize this to differential forms of higher degree unless there is a multiplication on the tangent spaces. For instance, on the tangent space of a Lie group one has the Lie bracket.

8. May 29, 2015

### "Don't panic!"

So does the relation $df(\mathbf{v})=\mathbf{v}(f)$ follow as a particular case of $\left(df_{p}(\mathbf{v})\right)(g)=\mathbf{v}_{p}(g\circ f)$ in which $g=\text{id}:\mathbb{R}\rightarrow\mathbb{R}$?

9. May 29, 2015

### lavinia

Yes. If f is real valued then v.f is a number. But at the same time df(v) is a tangent vector to R. The relation is that all tangent vectors to R are multiples of ∂/∂x and v.f is the multiple for df(v).

10. May 29, 2015

### Fredrik

Staff Emeritus
The pushforward $f_*:T_pM\to T_{f(p)}M$ is defined by $(f_*v)(g)=v(g\circ f)$. The $(\mathrm df)_p$ defined by $(\mathrm df)_p(v)=v(f)$ is a map from $T_pM$ into $\mathbb R$.

Edit: OK, I see now that there's more than one $df_p$ in post #1. The other one is the pushforward, as defined above.

Last edited: May 29, 2015
11. May 29, 2015

### "Don't panic!"

I think what's confused me in the past about the definition is that $df$ is by definition the map $df:T_{p}M\rightarrow\mathbb{R}$, but $\mathbf{v}$ is by definition the map $\mathbf{v}:\mathscr{F}(M)\rightarrow\mathbb{R}$ (where $\mathscr{F}(M)$ is the set of smooth functions from $M$ to $\mathbb{R}$) so I found it confusing how they could coincide, i.e. $df(\mathbf{v})=\mathbf{v}(f)$. The only way I could understand it was through applying a special case of the pushforward (differential) map between tangent spaces, but maybe I'm missing something?!

12. May 29, 2015

### Fredrik

Staff Emeritus
They coincide because df is defined to ensure that they do. I guess what you're really wondering is why it's defined that way. To think that this question has an answer, you must have some thoughts on what df is supposed to be. Maybe a small change in the value of f along the direction of v?

When $f:\mathbb R\to\mathbb R$, some books define $df:\mathbb R^2\to\mathbb R$ by $df(x,h)=f'(x)h$ for all x and all h. This makes df(x,h) approximately equal to f(x+h)-f(x), since we have
$$f'(x)\approx \frac{f(x+h)-f(x)}{h}.$$ Note that for all real numbers dx, we have df(x,dx)/dx=f'(x).

This idea has a fairly obvious generalization to the case where $f:\mathbb R^n\to\mathbb R$. Now $df:\mathbb R^n\times\mathbb R^n\to\mathbb R$ is defined by $df(x,h)=f_{,i}(x)h^i$ for all $x,h\in\mathbb R^n$.

When f is a function on a manifold and x is a coordinate system, we have
$$df(v)=v(f)=v^i\frac{\partial}{\partial x^i}\bigg|_p f =v^i (f\circ x^{-1})_{,i}(x(p)) \approx (f\circ x^{-1})(x(p)+\bar v)-(f\circ x^{-1})(x(p)),$$ where $\bar v$ is defined by $\bar v=(v^1,\dots,v^n)$. In the special case where the manifold is $\mathbb R^n$ and $x$ is the identity map, the above reduces to
$$df(v)=v^i f_{,i}(p)\approx f(p+\bar v)-f(p).$$ By the calculus definition of df, we have
$$df(p,\bar v)=v^i f_{,i}(p)\approx f(p+\bar v)-f(p).$$ So the differential geometry df can certainly be thought of as a generalization of the df in calculus.

13. May 30, 2015

### "Don't panic!"

So $\mathbf{v}$ acting on $f$ describes the differential (first-order) change in $f$ along the direction defined by $\mathbf{v}$. It is the directional derivative of $f$ along $\mathbf{v}$. Now, $df$ acting on $\mathbf{v}$ maps to a number quantifying the first-order change in $f$ as we "move" along $\mathbf{v}$. So the two descriptions match up.
However, I'm struggling to see how their definition in terms of mappings (given in my previous post) match up (it makes sense to me if we consider a pushforward of a vector to $T_{p}\mathbb{R}\cong\mathbb{R}$ acting on the identity map $\text{id}:\mathbb{R}\rightarrow\mathbb{R}$)?