Hyper $n-$ torus cohomology group?

Asked Dec 04 '14 at 00:46

Active Dec 04 '14 at 01:43

Viewed 86 times

I don't know if this interpretation is correct.

Is $S^{n_1} \times S^{n_2} \times \dots \times S^{n_k}$ some sort of hyper torus of dimension $1 + \sum_{k = 1} n_k$ (see here for calculation)?

Let's just take an example, what exactly is object $S^2 \times S^3$ (need a name for this set)? By Kinneth's formula (thanks in the comments), I found that

$H^p(S^2 \times S^3) = \oplus ( H^i(S^2) \otimes H^j(S^3) ) $. For $p = 0$, we get $\Bbb R \times \Bbb R = \Bbb R^2$. For $p \geq 1$, (I only did it up to $p = 2$), I found that it stays $\Bbb R^2$.

edited Apr 13 '17 at 12:21

Community

asked Dec 04 '14 at 00:46

Lemon

12,664

1

Since there is no definition of what a «hyper torus of dimension $m$» is, it is impossible to answer your first question. Your computation of $H^p(S^2\times S^3)$ is not correct, by the way. – Mariano Suárez-Álvarez Dec 04 '14 at 01:01
@MarianoSuárez-Alvarez, how is it now? – Lemon Dec 04 '14 at 01:23
Google for «Künneth formula» – Mariano Suárez-Álvarez Dec 04 '14 at 01:24
If you are using $|S^2\times S^3|$ to denote the dimension of $S^2\times S^3$, that is also wrong: that manifold has dimension 5. – Mariano Suárez-Álvarez Dec 04 '14 at 01:27
@MarianoSuárez-Alvarez, oh never heard of that formula before. So I guess by that formula, I would have $H^p(S^2 \times S^3) = \oplus H^i(S^2) \otimes H^j(S^3) $. I found that for $p = 0$, we get $\Bbb R \otimes \Bbb R = \Bbb R^2$. For $p \geq 1$, (I only did it up to $p = 2$), I found that it stays $\Bbb R^2$. – Lemon Dec 04 '14 at 01:42
But thank you anyways, I'll brutally work out the calculations myself. You helped me enough. – Lemon Dec 04 '14 at 03:34
$\mathbb{R}\otimes \mathbb{R}$ is actually just $\mathbb{R}$. – Kevin Carlson Dec 07 '14 at 02:37

Hyper $n-$ torus cohomology group?

0 Answers0