Understanding $(\mathbb{Z}/3)^2 \times_{\mathbb{Z}/3} M$

Question

I'm currently reading "Bordism of Elementary Abelian Groups via Inessential Brown-Peterson Homology" by Hanke (arXiv:1503.04563) and have come across some notation that I'm not familiar with. Let $\phi: \mathbb{Z}/3 \rightarrow (\mathbb{Z}/3)^2$ be a group homomorphism and $S^5$ the five dimensional sphere in $\mathbb{C}^4$. $\phi$ induces an action of $\mathbb{Z}/3$ on $(\mathbb{Z}/3)^2$ and $S^5$ comes with a natural $\mathbb{Z}/3$ action, the standard action from the lens space construction. The paper then writes $$(\mathbb{Z}/3)^2 \times_{\mathbb{Z}/3} S^5$$ How is this defined? Furthermore, it is stated that this comes with a $(\mathbb{Z}/3)^2$ action.

As a check, it should be defined so that $$((\mathbb{Z}/3)^2 \times_{(\mathbb{Z}/3)^2} (S^{2m_1+1}\times S^{2m_2+1})/(\mathbb{Z}_3)^2 \cong L^{2m_1+2}_3\times L^{2m_2+1}_3$$ is a product of lens spaces.

Answer 1

This has been more or less said in the comments, but it seems like someone should write an answer.

If $H$ is a subgroup of $G$ and $H$ has a left action on a space $X$, then a space $G\times _HX$ is defined, a quotient space of $G\times X$, by declaring $(gh,x)\sim (g,hx)$. Note that we are using the right action of $H$ on the space $G$. (If we like, we can create a left action on $G\times X$ using the right action on $G$ and the left action on $X$, writing $h(g,x)= (gh^{-1},hx)$. But this is unnecessary.)

Let's denote the equivalence class of $(g,x)$ by, say, $[g,x]$. The quotient space gets a left $G$-action, defined by $g_1[g_2,x]=[g_1g_2,x]$. (This is well-defined because the left action of $G$ on (the space) $G$ commutes with the right action: $[g_1(g_2h),x]=[(g_1g_2)h,x]=[g_1g_2,hx]$.

In categorical terms: The subgroup $H\subset G$ (or the homomorphism $H\to G$) is giving us a functor $X\mapsto G\times_HX$ from the category of spaces with left $H$-action to the category of spaces with left $G$-action. This functor is left adjoint to the forgetful functor.

This is very similar to how a homomorphism $R\to S$ of (associative) rings gives a functor $X\mapsto S\otimes_R X$ from $R$-modules to $S$-modules, again left adjoint to the forgetful functor.