1

I am reading the book "Differentiable Algebraic Topology, From Statifolds to Exotic Spheres" by Mathias Kreck.

In the page 136, he said:

We consider the flip diffeomorphism $\tau:M\times N\longrightarrow N\times M$ mapping $(x,y)$ to $(y,x)$, when $M$ and $N$ are oriented manifolds with dimensions $m$ and $n$, respectively. Then, $\tau$ changes the orientation by $(-1)^{mn}$.

However, in the whole book, he never mentioned this $\tau$ again. I believe this is related to a post here: Does the orientation on a product of manifolds depend on the order of the product?. However, this post did not provide the explicit formula of such a $\tau$.

A similar proof to vector space can be found here: $I(X,Z)=(-1)^{(\dim X) (\dim Z)}I(Z,X)$, and one could see that this diffeomorphism is the change of basis matrix.

However, do we have similar thing in manifolds? Can we modify the proof in the second link, to argue with the change of basis matrices?

Thank you!

Edit 1: (Question Solved)

As Tyrone pointed out, the orientation of a manifold is inherited from its corresponding tangent bundle. The orientation of the tangent bundle is the orientation of the vector space to which the fiber bundle isomorphic to (think about the trivialization map in the notion of vector bundle).

Then, the proof is immediate from the proof of the case of vector space, since we can treat the tangent space at each point as a vector space, and we have $T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N,$ as vector spaces.

I did not know this fact pointed out by Tyrone, since my geometry course only talked about when the tangent bundle (or manifold) is orientable, without mentioning where the orientation comes from. I read some books for this, but it involves some $2-$fold covering space, section map and the notion of orientation character, homology, etc.

I understand these basic notions, but I don't want to go too deep for my question, since it seems that my confusion does not need these many notions to be clarified.

I have written a proof in the answer of my own question, but it is really appreciated if anyone have another, either similarly shorter or more intrinsically inspiring answer.

I also sincerely appreciate all the discussions and helps from comments.

JacobsonRadical
  • 5,810
  • When you write $M \times N = (-1)^{\cdots}N \times M$, what do you mean? How are you multiplying $N \times M$ by $\pm 1$? – Rob Arthan Apr 07 '20 at 20:43
  • @RobArthan If $M$ has an orientation, I denote $-M$ to be the inverse orientation. What I wrote there means the orientation of $M\times N$ is $(-1)^{\cdots}$ times the orientation of $N\times M$. – JacobsonRadical Apr 07 '20 at 21:02
  • Yes, but those are orientations on two different manifolds. What I think you mean that there is a function that maps pairs comprising an orientation on $M$ and an orientation on $N$ to an orientation on $M \times N$ that respects the natural homeomorphism between $M \times N$ and $N \times M$ modulo the given factor of the form $\pm 1$. – Rob Arthan Apr 07 '20 at 21:12
  • @RobArthan yes. I think your clarification about the fact that they are two different manifolds is essential. and yes, I am asking for what you mentioned. Thank you so much. – JacobsonRadical Apr 07 '20 at 21:23
  • @RobArthan I am gonna edit my post. – JacobsonRadical Apr 07 '20 at 21:51
  • 1
    If you understand an 'orientation' of a manifold to be a coherent orientation of every tangent space (as a vector space), then exactly the same notions as those from linear algebra apply. – Tyrone Apr 08 '20 at 00:21
  • @Tyrone so the orientation of a manifold is inherited from its tangent bundle? which is locally (the fiber) is a vector space. I did not know this since it seems that my geometry course only talked about when a manifold or a tangent bundle is orientable, without talking where the orientation comes from, – JacobsonRadical Apr 08 '20 at 14:18
  • Yes. There are a couple of equivalent definitions. One is that an orientation is a covering of $M$ with coordinate charts such that the linear maps induced by the change of charts are orientation preserving (when you give $\mathbb{R}^n$ your favourite orientation). See Lee's book Introduction to Smooth Manifolds Chapter 15, pg. 380. A manifold is orientable when it admits an orientation. Not every manifold is orientable ($\mathbb{R}P^{2n}$, for instance), but this is generally easier to show using other definitions (say, by homology - see Lee Chapter 17, Theorems 17.31, 17.32). – Tyrone Apr 08 '20 at 15:06
  • @Tyrone thank you so much for this detailed clarification. Then the proof is fairly easy to me since I know how to prove the case of vector space, and the case tangent space is the same (if we treat them as vector space). – JacobsonRadical Apr 08 '20 at 15:09

1 Answers1

1

Okay, as I said in my edit, thanks to Tyrone, we need to only consider the tangent space at each point as a vector space, so the proof follows immediately from a proof of vector space. That is:

Lemma: For $W$ and $V$ two oriented vector spaces, in terms of orientation, we have $$V\oplus W=(-1)^{\dim V\dim W}(W\oplus V).$$

Proof of Lemma:

Suppose $\dim(V)=n$ and $\dim(W)=m$ so that $\dim(V\oplus W)=n+m$. Denote $Y:=V\oplus W$ and since $V$ and $W$ are oriented, we know that $Y$ must be oriented, and thus it has a positive ordered basis $B=(y_{1},\cdots, y_{n+m}).$

Since $V,W$ are oriented, they have positive ordered bases $B_{V}=(v_{1},\cdots, v_{n})$ and $B_{W}=(w_{1},\cdots, w_{m})$. We write $B_{V}B_{W}$ as the concatenation of $B_{V}$ and $B_{W}$, namely the sequence $(v_{1},\cdots, v_{n}, w_{1},\cdots, w_{m})$.

This clearly induces an ordered basis for $Y:=V\oplus W$. Let us denote the change of basis matrix between two basis $b_{1}, b_{2}$ to be $C(b_{1}, b_{2})$. Then to decide if the decomposition $V\oplus W$ of the oriented vector space $Y$ as a direct sum of two oriented subspaces is positive or negative, we need to know if $B_{V}B_{W}$ is a positive or negative ordered basis for $V$, that is, to decide if the change of basis $C(B, B_{V}B_{W})$ has positive or negative determinant.

Firstly, we know that $C(B_{V}B_{W}, B_{W}B_{V})=(-1)^{nm}$ and thus we have $$C(B, B_{V}B_{W})=C(B_{V}B_{W}, B_{W}B_{V})C(B, B_{W}B_{V})=(-1)^{nm}C(B, B_{W}B_{V}).$$

This immediately tells us the "sign" of the decomposition $V\oplus W$ of $Y$ is $(-1)^{nm}$ times the "sign" of the decomposition $W\oplus V$.

Hence, in terms of orientation, we have $$V\oplus W=(-1)^{\dim(V)\dim(W)}(W\oplus V).$$

Proof of my question:

We know that an orientation of a manifold is the coherent orientation of every tangent space of it as a vector space, so to know how the orientation of $M\times N$ is related to that of $N\times M$, we only need to know the relationship between $T_{(p,q)}(M\times N)$ and $T_{(q,p)}(N\times M)$ in terms of orientation.

But we know that $T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N$ and $T_{(q,p)}(N\times M)=T_{q}N\oplus T_{p}M$ as vector spaces, and thus it follows from the Lemma that $$T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N=(-1)^{\dim(M)\dim(N)}\Big(T_{q}N\oplus T_{p}M\Big)=(-1)^{\dim(M)\dim(N)}T_{(q,p)}(N\times M).$$

Therefore, in terms of orientation $$M\times N=(-1)^{\dim(M)\dim(N)}(N\times M).$$

Back to what I addressed in my question, the flip diffeomorphism is then the change of basis matrix between the basis of $Y:=T_{p}M\oplus T_{q}N$ and the concatenation of the basis $T_{p}M$ with $T_{q}N$, as what we did in the case of vector space.

JacobsonRadical
  • 5,810