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My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave.

As part of Proposition 11.13(i), I'm trying to compute the degree of the "interchanging" $T: J \times K \to K \times J, T(x,y)=(y,x)$, using the definition that for any $\omega \in \Omega^{d+l}(K \times J)$,

$$\deg(T)\int_{K \times J} \omega = \int_{J \times K} T^{*}\omega \tag{A}$$

The assumptions are:

  • $J$ and $K$ are compact, connected, oriented smooth submanifolds of $\mathbb R^{n+1}$ with respective dimensions $d$ and $l$ with $d+l=n$

  • $\tau \in \Omega^d(J)$ and $\sigma \in \Omega^l(K)$ give the respective orientations of $J$ and $K$

  • $K \times J$, also compact, connected oriented smooth submanifold of $\mathbb R^{n+1}$, specifically of dimension $d+l$, is given the product orientation $\gamma := \pi_K^{*}\sigma \wedge \pi_J^{*}\tau$, where $\pi_J: K \times J \to J, \pi_J(k,j)=j$ and $\pi_K: K \times J \to K, \pi_K(k,j)=k$ are projections.

  • Similarly, $J \times K$, also compact, connected oriented smooth submanifold of $\mathbb R^{n+1}$, specifically of dimension $d+l$, is given the product orientation $\psi := \pi_J'^{*}\tau \wedge \pi_K'^{*}\sigma$, where $\pi_J': J \times K \to J, \pi_J'(j,k)=j$ and $\pi_K': J \times K \to K, \pi_K'(j,k)=k$ are projections.

My idea is that in (A), I choose $\omega = \gamma$. I would pick the volume form like with the antipodal map (B), but I think any orientation form will do.

Firstly, $T^{*}\gamma = (-1)^{dl} \psi$ because

$$T^{*}\gamma = T^{*}(\pi_K^{*}\sigma \wedge \pi_J^{*}\tau)$$

$$= T^{*}(\pi_K^{*}\sigma) \wedge T^{*}(\pi_J^{*}\tau) = (T^{*}\pi_K)^{*}\sigma \wedge (T^{*}\pi_J)^{*}\tau$$

$$= (\pi_K')^{*}\sigma \wedge (\pi_J')^{*}\tau = (-1)^{dl} (\pi_J')^{*}\tau \wedge (\pi_K')^{*}\sigma = (-1)^{dl} \psi$$

because $\pi_K' = \pi_K \circ T$ and $\pi_J' = \pi_J \circ T$.

Secondly, $$\deg(T)\int_{K \times J} \gamma = \int_{J \times K} (-1)^{dl} \psi$$

Finally, $\deg(T) = (-1)^{dl}$ if I can somehow show $$\int_{K \times J} \gamma = \int_{J \times K} \psi \tag{C}$$

Question: How do we show (C), if (C) is correct? Otherwise, how do we approach computing $\deg T$?

Thoughts:

Intuitively, I think this is true by symmetry, but I'm not sure why precisely. I think there's some change of variables rule I'm missing here such as...

Perhaps we have something like:

$$\int_{K \times J} \gamma = \int_{T(J \times K)} \gamma = (-1)^{dl} \int_{J \times K} T^{*}\gamma = \int_{J \times K} (-1)^{dl} (-1)^{dl} \psi$$

because for some reason

$$\int_{T(J \times K)} \gamma = (-1)^{dl} \int_{J \times K} T^{*}\gamma$$

  • This might follow by integral a under diffeomorphism (Tu Problem 23.3), but I'm not exactly sure how this applies. Can I say that we take cases for $dl$ odd and $dl$ even to get $$\int_{T(J \times K)} \gamma = \pm \int_{J \times K} T^{*}\gamma$$ where $1=(-1)^{dl}$ for even $dl$ and $-1=(-1)^{dl}$ for odd $dl$?

(B) I'm not sure that the above link uses the definition of degree in (A). Without homotopy, I would say $$\deg(\text{Antipodal}) \int_{S^n} \text{vol}_{S^n} = \int_{S^n} \text{Antipodal}^{*}(\text{vol}_{S^n}) = \int_{S^n} (-1)^{n+1}\text{vol}_{S^n}$$

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    So how does the product orientation on $\Bbb R^k\times\Bbb R^\ell$ compare to the product orientation on $\Bbb R^\ell\times\Bbb R^k$? Think in terms of properties of wedge product. You should get a sign of $(-1)^{k\ell}$, as you suggest. I think you're just burying yourself in notation. Simplify! ... For your last question with $\gamma$ and $\psi$, think Fubini. – Ted Shifrin May 08 '19 at 23:24
  • @TedShifrin Thanks. Even if we were to consider those product orientations, I think we still have to do some change of variables like $T^{*}(1_{J \times K}) = 1_{K \times J}$. My thought is $$\int_{J \times K} \psi = \int_{\mathbb R^k \times \mathbb R^l} \psi 1_{J \times K} = \int_{\mathbb R^k \times \mathbb R^l} \alpha|{J \times K} 1{J \times K} = \int_{\mathbb R^k \times \mathbb R^l} \alpha 1_{J \times K} = \int_{J \times K} \alpha$$ where $\alpha$ is product orientation on $\mathbb R^k \times \mathbb R^l$. Then... –  May 09 '19 at 01:15
  • $$\int_{J \times K} \alpha = \int_{T^{-1}(K \times J)} \alpha = (-1)^{kl}\int_{K \times J} (T^{-1})^{*}\alpha = (-1)^{kl}\int_{K \times J} (-1)^{kl} \beta$$ where $\beta$ is product orientation on $\mathbb R^l \times \mathbb R^k$ (Actually $d$ not $k$). Am I wrong? –  May 09 '19 at 01:19

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This turns out to be unnecessarily confusing. :) If we fix an orientation on $\Bbb R^{d+\ell} = \Bbb R^d\times\Bbb R^\ell = \Bbb R^\ell\times\Bbb R^d$, then the map $T\colon\Bbb R^{d+\ell}\to\Bbb R^{k+\ell}$ given by $T(x,y)=(y,x)$ indeed has degree $(-1)^{d\ell}$. But in fact this sign persists.

For the moment, I'm going to use $T$ for tangent space and so I'll call the switching map $S$ instead. Specifically, if $e_1,\dots,e_d$ is an oriented basis for $T_xJ$ and $f_1,\dots,f_\ell$ is an oriented basis for $T_yK$, then $(e_1,0),\dots,(e_d,0),(0,f_1),\dots,(0,f_\ell)$ gives an oriented basis for $T_{(x,y)}(J\times K)$ and $(f_1,0),\dots,(f_\ell,0),(0,e_1),\dots,(0,e_d)$ gives an oriented basis for $T_{(y,x)}(K\times J$. On the other hand, $S_*(e_i,0)=(0,e_i)$ and $S_*(0,f_j)=(f_j,0)$. The oriented basis $(0,e_1),\dots,(0,e_d),(f_1,0),\dots,(f_\ell,0)$ indeed differs from the positive basis by a sign of $(-1)^{d\ell}$.

This amounts to the same computation you did showing that $T^*\gamma = (-1)^{d\ell}\psi$.

Let me comment that when it comes to integrating and applying Fubini's Theorem, to convert the integral of an $n$-form over a rectangle, say, in $\Bbb R^k$ into a $k$-fold multiple integral, you must first write the $n$-form as $f\,dx^1\wedge\dots\wedge dx^k$, where $dx^1\wedge\dots\wedge dx^k$ is a positively-oriented $k$-form according to the orientation on the rectangle. Then one integrates $f$ and uses Fubini. In our case, we must be sure to use $\gamma$ and $\psi$ as the respective "basis" $k$-forms.

Ted Shifrin
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