5
$\begingroup$

I am taking a course on smooth manifolds. We have just defined a tangent space of a manifold $M$ at a point $p\in M$ using paths $\gamma:\mathbb{R}\to M$, $\gamma(0)=p$. So $T_pM:=\{\gamma:\mathbb{R}\to M\;|\;\gamma(0)=p\}/\sim$, where $\sim$ is the equivalence relation of two paths having the same speed. That is, for two paths $\gamma_1,\gamma_2$ and a chart $\phi$, $$\left.\frac{d}{dt}\phi\circ\gamma_1\right|_{t=0}=\left.\frac{d}{dt}\phi\circ\gamma_2\right|_{t=0}\iff\text{$\gamma_1$ and $\gamma_2$ have the same speed.}$$ We defined addition and scalar multiplication of paths as follows: $$[\gamma_1]+\lambda[\gamma_2]:=[\phi^{-1}(\phi\circ\gamma_1+\lambda\phi\circ\gamma_2)],$$ where $\phi$ is a chart.
Now there's an exercise that I really don't know how to do. Given a smooth function $f:\mathbb{R}^m\to\mathbb{R}^n$, compute $f_*:T_p\mathbb{R}^m\to T_{f(p)}\mathbb{R}^n$ in the basis given by the coordinate vectors. Here $f_*$ is the function defined by $f_*:[\gamma]\mapsto[f\circ\gamma]$.
I am not even sure where to start. I have already shown that $f_*$ is linear.

EDIT:
I think this works. Write $f=(f_1,\ldots,f_n)$ where $f_i:\mathbb{R}^m\to\mathbb{R}$ and let $\gamma:\mathbb{R}\to\mathbb{R}^m$ be a path with $\gamma(0)=p=(p_1,\ldots,p_m)\in\mathbb{R}^m$. Let $\tilde{\gamma}_i(t):=(p_1,\ldots,p_{i-1},p_i+t,p_{i+1},\ldots,p_m)$ be the path along the $i$th basis vector in $\mathbb{R}^m$ through $p$. Then we have $\lambda_i\in\mathbb{R}$ such that $\gamma=\sum_{i=1}^m\lambda_i\tilde{\gamma}_i$. We have $$\begin{aligned} f_*[\gamma]&=f_*\left[\sum_{i=1}^n\lambda_i\tilde{\gamma}_i\right]\\ &=\sum_{i=1}^m\lambda_if_*[\tilde{\gamma}_i]\\ &=\sum_{i=1}^m\lambda_i[f\circ\tilde{\gamma}_i]\\ &=\sum_{i=1}^m\lambda_i\left.\frac{d}{dt}f\circ\tilde{\gamma}_i(t)\right|_{t=0}\\ &=\sum_{i=1}^m\lambda_iDf(\tilde{\gamma}_i(0))\left.\frac{d}{dt}\tilde{\gamma}_i(t)\right|_{t=0}\\ &=\sum_{i=1}^m\lambda_iDf(p)\left.\frac{d}{dt}\tilde{\gamma}_i(t)\right|_{t=0}\\ &=Df(p)\sum_{i=1}^m\left.\frac{d}{dt}\lambda_i\tilde{\gamma}_i\right|_{t=0}\\ &=Df(p)[\gamma]. \end{aligned}$$ So this yields that the matrix of $f_*$ is $Df$ at $p$.
Is this correct?

$\endgroup$
  • $\begingroup$ Have you already considered what $T_p\mathbb{R}^m$ looks like, given any $p\in\mathbb{R}^m$? $\endgroup$ – HSN Sep 25 '14 at 9:21
2
$\begingroup$

First, $\sim$ is the equivalence class of paths where the paths have the same velocity at time $t = 0$, not the same speed.

Since $f_*$ is linear, you know it's enough to compute $f$ for the given basis of coordinate vectors, which I'll denote $(\partial_{x^i})_p$. Moreover, by composing with translations we may as well align our charts so that $p = 0 \in \mathbb{R}^m$ and $f(0) = 0 \in \mathbb{R}^n$. This isn't strictly necessary, but given our definition of path addition it makes notation much cleaner.

Now, by definition, the $i$th coordinate vector $(\partial_{x^i})_0 \in T_0 \mathbb{R}^m$ at the point $0 \in \mathbb{R}^m$ is the equivalence class containing the curve $\gamma_i$ such that (1) is at $0$ at time zero (i.e., $\gamma_i(0) = 0$), and moves in the $i$th coordinate direction with unit speed, namely, $$\gamma_i(t) = (0, \ldots, 0, t, 0, \ldots, 0),$$ where $t$ is in the $i$th slot.

Now, we can compute the curve $f \circ \gamma_i$ in $\mathbb{R}^n$ that represents the pushforward $f_* [\gamma_i] = [f \circ \gamma_i]$. If we write $f$ in terms of its component functions, so that $$f(x^1, \ldots, x^m) = (f_1(x^1, \ldots, x^m), \ldots, f_n(x^1, \ldots, x^m)),$$ its $j$th component is $$f_j(x^1, \ldots, x^m),$$ and the $j$th component $(f \circ \gamma_i)_j = f_j \circ \gamma_i$ is $$f_j(0, \ldots, 0, t, 0, \ldots, 0).$$

The initial velocity $\left.\frac{d}{dt}\right\vert_0(f \circ \gamma_i)$ has $j$th component (in the coordinates $(y^j)$ on $\mathbb{R}^n$) is $$\left.\frac{d}{dt}\right\vert_0(f \circ \gamma_i)_j = \left.\frac{d}{dt}\right\vert_0(f \circ \gamma_i)_j = \left.\frac{d}{dt}\right\vert_0 f_j(0, \ldots, 0, t, 0, \ldots, 0),$$ which by the chain rule is just $$\frac{\partial f_j}{\partial x^i}(0),$$ and so the initial velocity of $f \circ \gamma_i$ is $$\left(\frac{\partial f_1}{\partial x^i}(0), \ldots, \frac{\partial f_n}{\partial x^i}(0)\right) \qquad (\ast)$$

Now, to compute the pushforward with respect to the coordinate vector basis $(\partial_{y^j})$, we must find curves that represent those coordinate vectors, and then decompose some representative of $[f \circ \gamma_i]$ as a linear combination of those curves. As before, we can pick as representative of each $\partial_{y^j}$ the curve $$\theta_j(t) := (0, \ldots, 0, t, 0, \ldots, 0),$$ where $t$ is in the $j$th slot.

It's now easy to write down a (unique) linear combination of the $\theta_j$'s in $[f \circ \gamma_i]$, that is, that has initial velocity $(\ast)$, because we already know the components w.r.t. to the coordinate basis, namely $$\sum_{j = 1}^n \frac{\partial f_j}{\partial x^i}(0) \theta_j,$$ and passing to equivalence classes gives that $$f_* [\gamma_i] = [f \circ \gamma_i] = \left[\sum_{j = 1}^n \frac{\partial f_j}{\partial x^i}(0) \theta_j\right] = \sum_{j = 1}^n \frac{\partial f_j}{\partial x^i}(0) [\theta_j].$$

So, the $(j, i)$ entry of the matrix representation of the pushforward $f_*: T_0 \mathbb{R}^m \to T_0 \mathbb{R}^n$, with respect to the bases $(\partial_{x^i}) = ([\gamma_i])$ of $T_0 \mathbb{R}^m$ and $(\partial_{y^j}) = ([\theta_j])$ of $T_0 \mathbb{R}^n$, is the partial derivative $\frac{\partial f_j}{\partial x^i}(0)$. But by definition this matrix is precisely the Jacobian $(Df)(0)$.

As an aside, in the setting of differential geometry, I actually prefer the functorial notation $T_p f$ instead of $f_*$ (or, as this exercise justifies, $(Df)(p)$, so that the pushforward of the map $f: M \to N$ at $p \in M$ is $T_p f: T_p M \to T_{f(p)} N$, which in particular emphasizes that the pushforward really is a map at a point.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, a few times I erased things I realized I didn't want to say (and your answer came out just as well I think). It's good to revisit the definitions of vectors from time to time, I think, since most of them are operationally pretty far from how they are actually used in practice. $\endgroup$ – Travis Willse Sep 25 '14 at 9:43
  • $\begingroup$ Thanks for your help! I have edited my question with my solution. Would you please check it for me? $\endgroup$ – Joffysloffy Sep 26 '14 at 16:02
  • $\begingroup$ You're welcome, I hope this is useful for you. In the fourth equality, you replace a vector (equivalence class) with the value of the time derivative, and we don't have a reason to equate these quantities (not yet, anyway, though you'll probably very soon learn you can compute them). Using the definitions at hand, you'll want to compute $[f \circ \gamma]$, find a nice path in $\mathbb{R}^n$ that has the same speed, and then write that path as a linear combination of the coordinate vectors at $T_{f(p)} \mathbb{R}^n$. (In the end, the matrix of $f_*$ turns out to be the Jacobian of $f$.) $\endgroup$ – Travis Willse Sep 26 '14 at 16:32
  • 1
    $\begingroup$ @Joffysloffy I added the rest of an argument, which in particular is very explicit about decomposing a representative of $[f \circ \gamma_i]$ in terms of a basis of $T_{f(p)} \mathbb{R}^n$. For convenience I've assumed we've translated (which doesn't affect the pushforward computation) our charts so that $p = 0$ and $f(p) = 0$. This isn't strictly necessary, but it makes dealing with writing down linear combinations of curves using our curve addition rules much cleaner. $\endgroup$ – Travis Willse Sep 27 '14 at 7:39
  • 1
    $\begingroup$ @Joffysloffy You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Willse Sep 28 '14 at 3:35
2
$\begingroup$

Let $d_{1,p},d_{2,p},\dots,d_{m,p}$ be paths on $\mathbb{R}^n$ such that $d_{i,p}(0)=p$ and $d^\prime_{i,p}(0)=e_i$ where $e_i=(0,\dots,0,1,0,\dots,0)\in\mathbb{R}^m$ is the $i$th unit vector.

Then $([d_{i,p}])_{i=1,\dots,m}$ forms a basis of $T_p\mathbb{R}^m$.

You should think of $[d_{i,p}]\in T_p\mathbb{R}^m$ as the partial derivative in the $i$th coordinate direction based in the point $p$.

Now the job is to compute $f_*$ in terms of this basis. That is, compute the coefficients in the expansion of $f_*([d_{i,p}])$ in terms of the basis $([d_{j,f(p)}])_{j=1,\dots,n}$.

So we start

$$f_*([d_{i,p}])=[f\circ d_{i,p}]\in T_{f(p)}\mathbb{R}^n$$

We now need to compute $\frac{d}{dt} (f\circ d_{i,p})(t)\left.\right|_{t=0}$, since the $j$th component of the velocity vector of the path at $0$ will be exactly the coefficient of $[d_{j,f(p)}]$. Use the chain rule and the definition of $d_{i,p}$!

You will see that the resulting matrix of coefficients representing $f_*$ is simply the Jacobian of $f$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot! This was very helpful! Unfortunately I can only accept one answer. $\endgroup$ – Joffysloffy Sep 26 '14 at 16:03
0
$\begingroup$

Heh, here you just need to compute.
Let $\gamma$ be a curve on $\mathbb{R}^n$, what does $\frac{d}{dt}f(\gamma)$ look like? Try using the chain rule.
If it helps, try this simple example, $f:\mathbb{R}^2 \to \mathbb{R}^3;(x,y)\to(x,y,x^2+y^2)$. Compute the pushforward of the curves $\gamma_1(t) = (cos(t),sin(t))$, $\gamma_2(t) = (1,t)$, $\gamma_3(t) = (1,cos(t))$ so you can see what's going on.
As a worked example take $\gamma(t) = (t,t)$.
$\frac{d}{dt}f(\gamma(t)) = \frac{d}{dt} (t,t,2t^2) = (1,1,2t)$.

Now, let us look at a slightly more general example, $\gamma$ is just some curve.
$\frac{d}{dt}f(\gamma(t)) = \frac{d}{dt} (\gamma_x,\gamma_y,\gamma_x^2 +\gamma_y^2) =(\gamma'_x,\gamma'_y,\gamma'_x\gamma_x+\gamma'_y\gamma_y)$
We can translate this into $\frac{\partial}{\partial x}$ as so (remember that $\gamma_x$ is just the x component of the curve;
$(\gamma'_x,\gamma'_y,\gamma'_x\gamma_x+\gamma'_y\gamma_y) = (1+\gamma_x)\frac{\partial}{\partial x} + (1+\gamma_y)\frac{\partial}{\partial y}$

Has this cleared things up for you? If you still need help I'll work through the original problem for you but I think it's usually better to struggle with these things when you start off. As they say, you only learn by making mistakes!

If you'd like a supplement I'd recommend An Introduction to Manifolds by Loring Tu or An Introduction to Smooth Manifolds by John Lee.

| cite | improve this answer | |
$\endgroup$

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.