6

I'm trying to show that the maps defined in https://math.stackexchange.com/a/413846/500094 are mutual inverses. I have problem with one direction:

$(g\circ f)(v)=g(d(\pi_X)_{(p,q)}v,d(\pi_Y)_{(p,q)}v)=d(\iota_X)_p(d(\pi_X)_{(p,q)}v)+d(\iota_Y)_q(d(\pi_X)_{(p,q)}v)=\\d(\iota_X\circ\pi_X)_{(p,q)}v+d(\iota_Y\circ\pi_Y)_{(p,q)}v=v+v=2v\ne v$

At which point am I mistaken?

Minato
  • 1,541
user557
  • 12,359
  • It's not true that $d(\iota_X\circ\pi_X){(p,q)}$ is the identity. (Though, it is true that $d(\pi_X \circ \iota_X)$ is!). It can't be the identity, since, by the chain rule, $d(\iota_X\circ \pi_X) = d\iota_X \circ d \pi_X$, so the map from $T{p,q} X\times Y$ to itself factors through the smaller dimensional $T_{p}X$. More concretely, if $v$ is tangent to ${p}\times Y$, then $d(\iota_X \circ \pi_X)) v = 0$ since already $d\pi_X(v) = 0$. – Jason DeVito - on hiatus Jan 22 '18 at 20:32
  • $d(\iota_X\circ \pi_X){(p,q)}$ is projection onto the subspace $T{(p,q)}(X\times {q})$. This might help. – AnonymousCoward Jan 22 '18 at 20:35
  • Geometrically, at any given point $(p,q)$ you can write any tangent vector uniquely as a sum of vectors orthogonal to $p\times Y$ and orthogonal to $X\times q$. The inverse simply takes two tangent vectors on $X$ at $p$ and on $Y$ at $q$ and sends them to the sum of these two orthogonal component vectors. The first observation says this is an isomorphism. – Pedro Jan 22 '18 at 20:47

1 Answers1

5

In general, if $W$ is a f.d. vector space and $P_V\colon W \to W$ is a projection map with image $V\subset W$ and kernel $U$, then $W \cong U \oplus V$ and the map $P_U+P_V = \text{id}_W$ (where $P_U\colon W \to W$ is the projection map with image $U$).

In this case, take $L = d(\iota_X\circ \pi_X)_{(p,q)}$.

  1. Show $L$ is a projection map ($L\circ L = L$).

  2. What is the image of $L$?

  3. What is the kernel of $L$?

  4. What map appearing in your question above is a formula for projection onto the kernel of $L$?

If you can answer these questions, you will have a nice answer to the problem.

Update: here are some extra details:

  1. We can use properties of pushforward and the fact that $\pi_X\circ \iota_X = \text{id}_X$ to show that $$ L\circ L = d(\iota_X\circ \pi_X)_{(p,q)}\circ d(\iota_X\circ \pi_X)_{(p,q)} = d(\iota_X\circ (\pi_X\circ \iota_X)\circ \pi_X)_{(p,q)} = d(\iota_X\circ \pi_X)_{(p,q)} = L$$ so $L$ is a projection map.

  2. The image of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the subspace $V = T_{(p,q)}X\times \{q\}$, the tangent space at $(p,q)$ of the submanifold $X\times \{q\}$. In short, the image of this map is the set of tangent vectors at $(p,q)$ to paths of the form: $$ \iota_X\circ \pi_x\circ \gamma: \mathbb{R} \to X\times Y$$ where $\gamma$ is a path in $X\times Y$ with $\gamma(0) = (p,q)$. Such a $\gamma$ can be written as $\gamma(t) = (\gamma_1(t),\gamma_2(t))$ for $\gamma_1\colon \mathbb{R}\to X$ and $\gamma_2\colon \mathbb{R}\to Y$. You should verify that $$\iota_X\circ \pi_x\circ \gamma(t) = (\gamma_1(t),q)$$ Thus the image of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the set of vectors in $T_{(p,q)}X\times Y$ that can be written in the form $$d(\iota_X\circ \pi_X)_{(p,q)}\gamma'(0) = \frac{d}{dt}|_{t=0}(\gamma_1(t),q) = (\gamma_1'(0),0)$$ which is the desired tangent space.

  3. The kernel of $d(\iota_X\circ \pi_X)_{(p,q)}$ is the subspace $U = T_{(p,q)}\{p\}\times Y$. I leave the justification to you.

  4. So which map above gives projection onto $U$? It's $d(\iota_Y\circ \pi_Y)_{(p,q)}$. The proof is similar to items 1. and 2. above. In particular, this map is NOT the identity map.
AnonymousCoward
  • 5,734
  • I've spoiled the answer to the second bullet point in the comments above, but you should still try to prove the first bullet point. – AnonymousCoward Jan 22 '18 at 20:52
  • Thank you. For now, having verified that $d(\iota_X \circ \pi_X){(p,q)}$ is not the identity, tried to fix my proof directly, and arrived at another contradiction. The differential $d(\iota_X \circ \pi_X){(p,q)}$ is the map that sends $v=(v_1,v_2)$ to $(v_1,0)$, right? At the same time the second differential sends $v$ to $v$. So their sum sends $v$ to $(2v_1,v_2)$, which should not be the case... – user557 Jan 22 '18 at 23:27
  • Let me expand the answer a bit for you. – AnonymousCoward Jan 23 '18 at 07:35
  • Isn't there a problem with the equation $d(\iota_X\circ \pi_X){(p,q)}\gamma'(0) = \frac{d}{dt}|{t=0}(\gamma_1(t),q) = (\gamma_1'(0),0)$? You seem to be using the same symbol for different objects and/or making some identification. Namely, the $\gamma'(0)$ on the LHS is a tangent vector in $T_{(p,q)}(X\times Y$), whereas the RHS is not a tangent vector. The derivative does not even make sense unless you are working in some coordinate chart, but even in this case, it is not clear to me what identification(s) you are using. (cont. below)... – Matematleta Aug 28 '19 at 05:41
  • .... The most you can say is that for suitable $f:X\times Y\to \mathbb R,\ d(\iota_X\circ \pi_X){(p,q)}\gamma'(0)(f) = \frac{d}{dt}|{t=0}f(\gamma_1(t),q).$ I am not claiming that the result is false, only that there is some explanation that I am missing here. – Matematleta Aug 28 '19 at 05:41
  • @Matematleta I don't know if I follow, is your comment simply about the question of defining tangent spaces using derivations vs. derivatives at zero of paths through the point? – AnonymousCoward Sep 04 '19 at 21:58
  • @Matematleta oh, is it simply that the expression "$(\gamma_1'(0),0)$" doesn't make sense? I think if you re-write it as the more complicated $\frac{d}{dt}|_{t=0} \iota_X\circ \pi_x\circ \gamma(t)$, which is I suppose what I meant, then it will make sense, and maybe something more needs to be said about the subspace spanned by those vectors that seemed totally obvious before. – AnonymousCoward Sep 04 '19 at 22:02
  • @AnonymousCoward yes I guess that's right. You seem to be treating derivations as ordinary derivatives, which is fine, as long as you are consistent. But the equation I cite does not make sense as it stands, as far as I can tell. I posted a full question here . I hope you can provide an answer. – Matematleta Sep 04 '19 at 22:07
  • @Matematleta it is my understanding that the definition of the tangent space via derivations is but one of two equivalent and equally valid definitions. The other being with tangent vectors. Both have their advantages. I think here I am pretty uniformly using the latter. Which is to say that I'm not "treating derivations as ordinary derivatives" since I'm not thinking at all in terms of derivations here. – AnonymousCoward Sep 04 '19 at 22:18