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Let us consider the torus $T = S^1_X \times S^1_Y$, where the former $S^1_X$ has the coordinate $X$, whereas the latter $S^1_Y$ has $Y$. We have an alternate property $dX \wedge dY = - dY \wedge dX \in H^2(T, {\Bbb Z})$.

I am still wondering what should be the similitude of this alternate property in $H_2(T, {\Bbb Z})$.

When we consider $[T] \in H_2(T, {\Bbb Z})$, like $H^2(T, {\Bbb Z})$ the order must be taken into account. More specifically, I would like to ask

Q. $\int_{S^1_X \times S^1_Y}dX \wedge dY = \int_{S^1_Y \times S^1_X}dX \wedge dY$?

It seems to me that $S^1_X \times S^1_Y$ is the minus of $S^1_Y \times S^1_X$ in $H_2(T, {\Bbb R})$, which is what I would like to ask in the previous question. I seem to miss some basic formalism in the duality of $H_2$ and $H^2$.

Does the above question make sense?

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    $\begingroup$ If you exchange $X$ and $Y$ in your RHS you will see that, indeed, there is a minus sign. Anyway, I'm not sure this is really a research-level question... $\endgroup$ Commented Jan 10, 2019 at 16:40
  • $\begingroup$ I'm sorry I cannot understand what you're asking. Since the torus is a commutative topological group, the Pontryagin product on $H_*(T;\mathbb{Z})$ is (skew-)commutative. There are no integrals required for this though. $\endgroup$ Commented Jan 10, 2019 at 16:41
  • $\begingroup$ $\int_{S^1_Y\times S^1_X}dX\wedge dY=\int_{S^1_X\times S^1_Y}dY\wedge dX=-\int_{S^1_X\times S^1_Y}dX\wedge dY \ \ $ — just change the names of the coordinates. This is far from being at research level. $\endgroup$
    – abx
    Commented Jan 10, 2019 at 16:58
  • $\begingroup$ Possible duplicate of Alternate property of H^2(T, Z) $\endgroup$
    – user44191
    Commented Jan 10, 2019 at 17:24

1 Answer 1

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In order to define integration of differential forms on a manifold $M$, one must choose an orientation of $M$. Strictly speaking, this means there isn't enough information to answer your question, but people often don't explicitly specify the orientation, and instead use whatever makes the most sense. The point is, I'm going to assume the standard orientations of $S_X^1\times S_Y^1$ and $S_Y^1\times S_X^1$, in which case they're oppositely oriented and you're correct that $[S_X^1\times S_Y^1] = -[S_Y^1\times S_X^1]\in H_2(T^2)$. But if you had a different orientation in mind, the answer could be different.

Suppose $V$ and $W$ are oriented vector spaces, that $(e_1,\dotsc,e_m)$ is a positively oriented basis of $V$, and that $(f_1,\dotsc,f_n)$ is a positively oriented basis of $W$. The standard orientation on $V\oplus W$ is the one for which the basis $(e_1,\dotsc,e_m,f_1,\dotsc,f_m)$ is positively oriented. This extends to the orientation convention for products of manifolds: if $M = N_1\times N_2$, then at any $(x_1,x_2)\in M$, $T_{(x_1,x_2)}M = T_{x_1}N_1\oplus T_{x_2}N_2$, so the convention defines an orientation $T_{(x_1,x_2)}M$ given orientations of $T_{x_1}N_1$ and $T_{x_2}N_2$.

The point is that the orientations on $N_1\times N_2$ and $N_2\times N_1$ may differ, and they do in this situation. The orientation on $S_X^1\times S_Y^1$ is the one in which the basis $(\partial_x, \partial_y)$ of $T_{(x,y)}T^2$ is positively oriented, and the orientation on $S_Y^1\times S_X^1$ is the one in which $(\partial_y, \partial_x)$ is positively oriented. The change-of-basis matrix between these bases is $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, which has negative determinant, so these two orientations are opposite. Therefore the answer to your question is no; in fact, $$\int_{S_X^1\times S_Y^1} \mathrm dX\wedge\mathrm dY = -\int_{S_Y^1\times S_X^1} \mathrm dX\wedge\mathrm dY.$$

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  • $\begingroup$ Great thanks! Pierre $\endgroup$
    – Pierre
    Commented Jan 10, 2019 at 21:25

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