# Degree of a map between product of manifolds

Let $M^m$ and $N^n$ be compact, oriented smooth manifolds without boundary. Then what is the degree of the map

$$f: M\times N \to N \times M$$

given by $f(x,y) = (y,x)$? I have the feeling it should be $(-1)^{mn}$ (would fit in with some proof I have to give), but the result I come up with is $1$.

My reasoning: If $w_1, \dots, w_m$ and $v_1, \dots, v_n$ are positively oriented ordered bases for $T_pM$ and $T_qN$, then $w_1, \dots, w_m, v_1, \dots, v_n$ is a positively oriented ordered basis on $T_{(p,q)}(M \times N)$ (where $T_{(p,q)}(M \times N) \simeq T_pM \oplus T_qN$).

Similarly, $v_1, \dots, v_n, w_1, \dots, w_m$ is a postiviely oriented basis for $T_{(q,p)}(N \times M)$. Now $df(p,q)(w_1, \dots, w_m, v_1, \dots, v_n) = (v_1, \dots, v_n, w_1, \dots, w_m)$, so $f$ is order preserving. And since it is a diffeomorphism, it's degree is $1$.

Now since the above would contradict another statement I have to prove, I must be making a mistake. I would be very glad if someone could help me out here. :-)

Best Regards,

S.L.

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## 2 Answers

When you say $df(p,q)(w_1,\ldots, w_m,v_1,\ldots, v_n) = (v_1,\ldots, v_n,w_1,\ldots, w_m)$, what exactly do you mean? My guess is that you're looking at the image of an ordered basis. I think it might be better if you look at each individual basis vector in $T_{(p,q)}(M\times N)$ and see what its image is instead of trying to take them all at once, so that you can be sure of what you're doing. Then, the question is not whether the map is order-preserving, but whether the image of your oriented basis has the right or wrong orientation. You can check this by checking the sign of the permutation that brings it back to your original choice of ordered basis.

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"... and see what its image is instead of trying to take them all at once, so that you can be sure of what you're doing". Very good advice, indeed. I'm such an id... :) Thanks a lot! – Sam Apr 1 '11 at 8:40
Sure! I've found that when I'm stuck, especially on differential geometry type stuff, usually my problems stem from not totally understanding the symbols I'm throwing around. I think it can be especially easy to fool yourself into thinking you understand something that you don't in this field. – Aaron Mazel-Gee Apr 2 '11 at 7:02

Since $f$ is one to one, then its degree is $-1$ or $1$ (also because $f\circ f=Id$, then $(\mathrm{deg}\,f)^2=1$).

if $\{e_1,...,e_{n+m}\}$ denote the canonical basis of $\mathbb{R}^{n+m}$ then $\det D(x,y)f=\det(e_{n+1},...,e_{n+m},e_1,...,e_n)=(-1)^{mn}\det(e_1,...,e_n,e_{n+1},...,e_{n+m})=(-1)^{mn}$.

So $\mathrm{deg}\,f=(-1)^{mn}$.

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