Alternate property of H^2(T, Z) [closed]

Question

Let us take $T = S^1_X \times S^1_Y$, which is a torus, where the former $S^1_X$ has the coordinate $X$, whereas the latter $S^1_Y$ has $Y$. If we consider the generator $dX \wedge dY \in H^2(T, {\Bbb R})$, we have obviously the well-known relation $dX \wedge dY = - dY \wedge dX$, i.e. an alternate property.

By dual, $H_2(T, {\Bbb R})$ must enjot the alternate property as well. This seems to consider $S^1_X \times S^1_Y$ or $S^1_Y \times S^1_X$.

Q. How $S^1_X \times S^1_Y$ is the minus of $S^1_Y \times S^1_X$ in $H_2(T, {\Bbb R})$?

Obviosuly, it is $NOT$ easy to check this compared to visible differential forms.

Answer 1

When one passes from cohomology to homology, the wedge product of differential forms becomes the intersection product. That is, if $N_1$ and $N_2$ are submanifolds of a smooth manifold $M$, and $N_1$ and $N_2$ are transverse, the intersection product of the homology classes represented by $N_1$ and $N_2$ is the homology class of $N_1\cap N_2$, which is also a submanifold by transversality. However, unless one takes mod 2 homology, one must choose orientations on $M$, $N_1$ and $N_2$ and consider the homology class of $N_1\cap N_2$ as an oriented submanifold. The sign change you're asking about comes from these orientations; specifically, the two intersections will be oppositely oriented, and therefore will differ by a sign.

A quick note about the intersection product is that it reduces dimension: this is because Poincaré duality exchanges $H^2(T^2;\mathbb R)$ and $H_0(T^2;\mathbb R)$, and exchanges $H^1$ and $H_1$, so the cup product $H^1\times H^1\to H^2$ is sent to the intersection product $H_1\times H_1\to H_0$. This is why the dual to $\mathrm dX\wedge\mathrm dY$ is the homology class of $S_X^1\cap S_Y^1$ in $H_0$, not $S_X^1\times S_Y^1$ in $H_2$.

Now the computation. Let's orient $T^2$ in the usual way, so that $(\partial_X, \partial_Y)$ is a positively oriented basis of the tangent space at any point. $S_X^1$ and $S_Y^1$ intersect at a single point, and do so transversely. The orientation convention for an isolated intersection point $p$ of oriented transverse submanifolds $N_1$ and $N_2$ of an oriented manifold $M$ is that we count that point as $+1$ if the orientations on $T_pN_1\oplus T_pN_2 = T_pM$ agree, and as $-1$ if they disagree. In our case, when we consider $S_X^1\cap S_Y^1$, the orientation on $T_pS_X^1\oplus T_pS_Y^1$ is the usual orientation, which agrees with the ambient orientation on $T_pT^2$, so the intersection number is $+1$. For $S_Y^1\cap S_X^1$, we obtain the opposite orientation, for which $(\partial_Y, \partial_X)$ is positively oriented, and therefore the intersection number is $-1$. Letting brackets denote homology classes, this means $$[S_X^1\cap S_Y^1] = -[S_Y^1\cap S_X^1]\in H_0(T^2;\mathbb R).$$