0

How can I Compute the homotopy groups $\pi_n(\mathbb RP^2 \times \mathbb S^3)$ ?

Which theorems will help me in this? Should I use Kunneth theorem?

I have seen this post here Calculate $[\mathbb{S}^n, \mathbb{RP}^{n-1}] \cong \pi_n(\mathbb{RP}^{n-1})/\pi_1(\mathbb{RP}^{n-1})$ but I am not sure how will this help me, if it will, by any means. Any clarification will be greatly appreciated!

Brain
  • 1,003
  • 1
    Try to construct its universal covering space. My answer to another question shows that your problem is the same as computing homotopy groups of spheres. Kunneth theorem is not very useful in the current setting since you are interested in homotopy group instead of homology group. – Kevin.S Dec 20 '23 at 04:39
  • Thank you :) @Kevin.S – Brain Dec 20 '23 at 04:46
  • @MarianoSuárez-Álvarez how is that, could you clarify please? – Brain Dec 20 '23 at 05:27

1 Answers1

3

How can I Compute the homotopy groups $\pi_n(\mathbb RP^2 \times \mathbb S^3)$ ?

It is well known that $\pi_n(X\times Y)$ is isomorphic to $\pi_n(X)\times \pi_n(Y)$.

And so being able to calculate $\pi_n(\mathbb RP^2 \times \mathbb S^3)$ means being able to calculate $\pi_n(\mathbb S^3)$. But AFAIK this still is one of the biggest open problem in homotopy theory. Except for $\mathbb{S}^1$ every sphere has some homotopy groups which we don't know how to calculate. And whatever we known about them is highly complicated, one may say: mathematics of the highest order.

If someone gave you this as an exercise, then there's some major mistake here. You mention Kunneth, so perhaps you were meant to calculate homology instead of homotopy groups? If so, then Kunneth (which applies to homology, not homotopy) is a good approach. Homologies are often easier to calculate then homotopies, especially for such well behaving spaces (CW complex).

freakish
  • 42,851
  • what about the solution for the fundamental groups of these spaces given in the comments above? Does not this contradict that the fundamental group of the product is the product of the fundamental groups? – Brain Dec 20 '23 at 16:36
  • @Brain $\pi_1$ is fully calculable. And there are couple of other $n$ for which we are able to calculate $\pi_n$. But there is no general solution. And none of the comments imply there is. – freakish Dec 20 '23 at 16:43
  • Also $\pi_1(\mathbb{S}^n)=0$ for any $n>1$, so that simplifies alot. – freakish Dec 20 '23 at 17:02
  • Then why the multiplication with the first homotopy group of RP^2 for example does not give us zero since pi_1 (S^2) = 0? – Brain Dec 20 '23 at 17:34
  • 2
    @Brain because that is not like numerical zero, and Cartesian product is not multiplication. $\pi_1(X)$ is a set, when we say it is zero, we mean zero group, or trivial group, or more precisely $0={e}$ is the trivial group with a single element. Then $X\times{e}\simeq X$. Some people also use $\pi_1(X)=1$ notation, but it is all the same. We mostly use "$0$" symbol because it is the neutral object with respect to Cartesian product "$\times$" which is a commutative operation. – freakish Dec 20 '23 at 18:27