4
$\begingroup$

From what I understand, the Borel construction takes a $G$-space $X$ and produces a topological space $X\times_{G}\mathbf{E}G$―the homotopy quotient $X/\!\!/G$ of $X$ by $G$ in the $\infty$-category of spaces $\mathcal{S}$―satisfying $$\mathrm{H}^*(X\times_G\mathbf{E}G;R)\cong\mathrm{H}^*_G(X;R).$$ One partiularly important example is given by $\mathbf{B}G$, the Borel construction / homotopy quotient of the point: $\mathbf{B}G\simeq\mathbf{E}G\times_G*$.

Moving to the pointed setting, one has the pointed Borel construction, which takes a pointed $G$-space $X$ and returns $\mathbf{E}G_+\wedge_{G}X$, the homotopy quotient of $X$ by $G$ in the $\infty$-category of pointed spaces $\mathcal{S}_*$. Concretely, it is given by \begin{align*} \mathbf{E}G_+\wedge_{G}X &\overset{\mathrm{def}}{=} \frac{\mathbf{E}G\times_G X}{\mathbf{E}G\times_G*},\\ &\cong \frac{\mathbf{E}G\times_G X}{\mathbf{B}G}. \end{align*} Now, $\mathbf{E}G_+\wedge_G*\cong*$, rather than $\mathbf{B}G$. But while $*$ is the monoidal unit of $\mathcal{S}$, it is not that of $\mathcal{S}_*$, which is $S^{0}$. Hence it would be interesting to know:

Question. What is the pointed Borel construction $\mathbf{E}G_+\wedge_G S^0$ of the $0$-sphere? Is it related to $\mathbf{B}G_+$?

$\endgroup$
3
  • 2
    $\begingroup$ Your definition is wrong. It's the same as EG x_G X modulo EG x_G pt; the basepoint you have in the numerator gets collapsed when you take the smash product. Beyond that, this is a definition push, so I'm not sure where you run into trouble. Yes, EG_+ ^_G S^0 = BG_+. But this is not really interesting: (X_+ ^ Y_+) = (X ^ Y)_+ in general. Quotienting by G changes nothing... $\endgroup$
    – mme
    Commented Aug 23, 2021 at 9:28
  • $\begingroup$ This is a sidenote: for any $\infty$-category $\mathcal C$ and a $G$-equivariant object $X\colon BG\to\mathcal C$, the homotopy orbit (i.e. "homotopy quotient" in your post) $X_{hG}$ is simply the colimit taken over $BG$. $\endgroup$
    – Z. M
    Commented Aug 23, 2021 at 16:48
  • $\begingroup$ @mme Thanks! I edited the question and corrected the definition. $\endgroup$
    – Emily
    Commented Aug 23, 2021 at 21:12

1 Answer 1

3
$\begingroup$

Let's apply your definition (which I think has typos on the RHS - the two "+" subscripts on the EG should not be there I think). Let's model everything as topological spaces and do the calculation there. Let's model $S^0$ as the discrete set $\{p,q\}$ with basepoint $p$.

$$EG_+\wedge_GS^0:=\frac{EG\times_G\{p,q\}}{EG\times_G\{p\}}$$ $$\simeq \frac{EG\times_G\{p\}\coprod EG\times_G\{q\}}{EG\times_G\{p\}}$$ $$\simeq \frac{BG\coprod BG}{BG}$$ $$\simeq \text{pt} \coprod BG $$ $$\simeq BG_+$$

$\endgroup$
1
  • $\begingroup$ Thanks, kiran, this is perfect! $\endgroup$
    – Emily
    Commented Aug 23, 2021 at 21:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .