Differential map between tangent spaces

["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?

 

[Ben Niehoff]

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

 

["Don't panic!"]

The map you are calling dfpd f_p is also called the pushforward f∗f_*.

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})?

 

[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).

 

["Don't panic!"]

the exterior derivative of a function is just the differential, i.e. it is the one form whose value at p is df(p).

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?

 

[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, $df$.

 

[lavinia]

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?

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.

 

["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}?

 

[lavinia]

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}?

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).

 

[Fredrik]

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

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.

 

["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?!

 

[Fredrik]

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.

 

["Don't panic!"]

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?

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})?