Inverses in the homotopy classes of maps into $RP^{\infty}$

Question

One can define bilinear maps $\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^{2n-1}$ by considering the elements in $\mathbb{R}^n$ as polynomials and doing multiplication. This defines an $H$-space structure on $RP^{\infty}$ since elements of this space have zeros after a finite number of entries in homogeneous coordinates. Apparently, this defines a group structure on the set $[X, RP^{\infty}]$ of homotopy classes of maps from a space $X$ into $RP^{\infty}$. I am having a hard time seeing how inverses work in this group and I can't seem to find any information in books or online. Any thoughts would be much appreciated!

Answer 1

Not a solution, just a little long for a comment.

I think there is a theorem of Stasheff to the effect that a stricly associative (?) H-space $S$ whose set of path components $\pi_0S$ is a group (under the induced map $\mu_*$, where $\mu:S\times S\to S$ is multiplication), that such an H-space is homotopy equivalent (as an H-space) to a loop space, thus is a group object.

I don't know how this gives the inverse map here, but the hypothesis are certainly verified, as $\Bbb RP^{\infty}$ with your multiplication is a strictly associative (stricly commutative, stricly unital) path-connected H-space. Thus the homotopy sets $[X,\Bbb RP^{\infty}]$ are actually commutative groups (and not just monoids.)

It may or may not help to use the fact that $\Bbb RP^{\infty}$ is a $K(\Bbb Z/2\Bbb Z,1)$; one would have to relate the multiplicative structures.

I seem to remember this is mentioned early on in Adam's book ''Infinite Loop Spaces'', and probably in Stasheff's articles ''Homotopy Associativity for H-spaces I, II'' which are available online for free.

Maybe an answer?

It seems not too unreasonable (at first glance) that $f$ is its own homotopy inverse. For $f^2$ is (intuitively, this is by no means a proof), at every point $x\in X$, a line in $\Bbb R^{\infty}$ generated by a vector whose greatest index with non zero coefficient is carried by a positive coefficient ($a_n^2$), and one might hope for a ''linear homotopy'' from $f^2$ to the constant map equal to $\Bbb R\in\Bbb RP^{\infty}$.

I think the following is a proof :

In a topological group $G$, multiplication of $G$ induces the same map as concatenation of loops. It seems to me this is true in a strictly unital H-space. Let $a,b$ be loops based at $1=1_{\Bbb RP^{\infty}}=\Bbb R\in\Bbb RP^{\infty}$, and let $[1]$ be the constant loop at $1$. Then $b$ is homotopic (as a loop) to $[1]\star b$, and similarly $a$ is homotopic (as a loop) to $a\star [1]$, and thus $$\mu(a,b)\simeq\mu(a\star [1],[1]\star b)=a\star b$$ (where we have used the fact that $1$ is a strict unit) so that multiplication $\mu_*$ is equal to multiplication in $\pi_1(\Bbb RP^{\infty},1)$.

(Edit) More precisely, if $(X,e)$ is a pointed space equipped with a multiplication $\mu:X\times X\to X$ that satisfies that the two maps $X\to X$ defined by $x\mapsto\mu(x,e)$ and $x\mapsto\mu(e,x)$ are homotopic rel. $e$ to $\mathrm{id}_{X}$, then $\mu_*:\pi_1(X,e)\times\pi_1(X,e)\to\pi_1(X,e)$ is product of paths. (End of Edit)

This shows that $\mu(f,f)_*$ induces the trivial map on $\pi_1$, since $\pi_1(\Bbb RP^{\infty})=\Bbb Z/2$ (and on all other homotopy groups since they are trivial for projective space.)

When $X$ is a CW complex (or any space that is locally connected in the suitable sense), $\mu(f,f)$ lifts to a map $\widetilde{\mu(f,f)}:X\to\Bbb S^{\infty}$ (using the standard facts about covering spaces). The infinite sphere is contractible, and so $\widetilde{\mu(f,f)}$ is homotopic to a constant map, and so is its projection $\mu(f,f)$.

This shows that $f$ is its own homotopy inverse.