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Let $SO(3)$ be the $3\times 3$ orthogonal group.

Define a map $i\colon SO(3)\to SO(3)\times S^2$ by $A\mapsto (A,A^{-1} e_1)$, where $e_1$ is the unit normal vector $(1,0,0)\in S^2$.

Then, $i$ induces $i_*\colon H_3(SO(3))\to H_3(SO(3)\times S^2)=H_3(SO(3))\oplus H_1(SO(3))$. (The last equality follows from the Kunneth formula.)

What is the image of the generator $[SO(3)]\in H_3(SO(3))$ via $i_*$?

I think the answer is $(1,1)\in H_3(SO(3))\oplus H_1(SO(3))$.

Is this true?

Thanks in advance.

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What is your strategy to prove that the second map $H_3(SO(3)) \to H_1(SO(3))$ is the reduction mod 2 map? – mland Jan 7 '13 at 10:15
I don't have... – user55417 Jan 7 '13 at 17:35
up vote 1 down vote accepted

Edit The following is a greatly simplified form of the original answer. (See the edit history if you're interested).

I think the image is actually is $(1,0)$. The first coordinate is easy: considering the compostion $SO(3)\rightarrow SO(3)\times S^2\rightarrow SO(3)$ where the second map is the natural projection $\pi$, we see that $\pi_\ast i_\ast = Id_\ast$ on $H_3(SO(3)) \cong \mathbb{Z}$. Since $\pi$ is an isomorphism when restricted to $$H_3(SO(3))\oplus 0\subseteq H_3(SO(3))\oplus H_1(SO(3))\otimes H_2(S^2)$$ this implies that the first coordinate is a $1$.

The second coordinate $0$ takes some more work. The starting point is to use naturality of the reduction $$H_3(SO(3);\mathbb{Z})\otimes \mathbb{Z}/2 \rightarrow H_3(SO(3); \mathbb{Z}/2)$$ to see that $$i_\ast([SO(3)] = (1,0)\in H_3(SO(3)\times S^2)$$ iff $$ i_\ast [SO(3)] = (\overline{1},\overline{0}) \in H_3(SO(3)\times S^2; \mathbb{Z}/2)\cong \mathbb{Z}/2\oplus\mathbb{Z}/2. \, \, \, (\ast\ast)$$

To compute the second $i_\ast$ map, first consider the map $\phi:SO(3)\times SO(3) \rightarrow SO(3)\times S^2$ given by $\phi(A,B) = (A,B^{-1} v_1)$. The point is that $i = \phi\circ \Delta$ where $\Delta:SO(3)\rightarrow SO(3)\times SO(3)$ is the diagonal embedding. So, we can compute $i_\ast$ as $\phi_\ast \Delta_\ast$.

The map $\Delta_\ast$ is easy - it sends $[SO(3)]$ to $[SO(3)\otimes 1] + [1\otimes SO(3)]$, so we need only understand $\phi_\ast$ on $[SO(3)\otimes 1]$ and $[1\otimes SO(3)]$. But since $\phi$ is simply a product of maps, we see $\phi_\ast[SO(3)\otimes 1] = [SO(3)\otimes 1]$ and $\phi_\ast [1\otimes SO(3)] = 0$ since $H_3(S^2;\mathbb{Z}/2) = 0$.

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