5
$\begingroup$

Let $X$ be an $H-space$, and let a multiplication $,\cdot,$ be given, associative up to homotopy.

Let $\Omega X$ be the loopspace of $X$ based at the identity and let the multiplication $ \circ $ on the loopspace be given by concatenating loops.

We also have a multiplication on the loopspace given by pointwise multiplication of loops. Since the loops are based at the identity this also give a loop based at the identity. Call this multiplication $\mu$.

Are the multiplication maps $\mu$, $\circ$ homotopic?

I am trying to verify the assertion in Odd primary exponents of Moore Spaces that if the power map $\dot k: X \to X$ sending $x$ to a choice of the $k$th power $x^k=(x(x(x(x...))$ $k$ times, is null homotopic, then the map $k^\circ$ sending $\gamma \in \Omega X$ to its kth homotopy exponent is also null homotopic.

Define $k^\mu$ to be the $k$-th power map on the loopspace using pointwise multiplication of loops.

Now if $\mu$ and $\circ$ are indeed homotopic, then since $k^\mu$ is null homotopic, it follows that $k^\circ$ is also null homotopic.

So is it true that $\mu$ and $\circ$ are homotopic?

$\endgroup$
3
  • $\begingroup$ I know it is true for the $H-space$ multiplication on $S^1$. I am trying to work this out for the multiplication on the $SU(2)$ case. $\endgroup$ Feb 9, 2016 at 14:41
  • $\begingroup$ This may come down to an Eckmann-Hilton arguement. $\endgroup$
    – Dan Rust
    Feb 9, 2016 at 15:52
  • 1
    $\begingroup$ Related. The question asks for a topological group, but it works exactly the same for a $H$-space. $\endgroup$ Jul 15, 2016 at 7:12

2 Answers 2

3
$\begingroup$

This follows by an Eckmann-Hilton argument and the Yoneda lemma. For any pointed space $Y$, $\mu$ and $\circ$ each give a binary operation on the set $[Y,\Omega X]$ of homotopy classes of pointed maps $Y\to\Omega X$. By Yoneda, $\mu$ and $\circ$ are homotopic iff these binary operations are equal for all $Y$.

Now observe that for any $Y$, the two binary operations on $[Y,\Omega X]$ both have the constant map as an identity, and they satisfy the interchange law $\mu(a\circ b,c\circ d)=\mu(a,c)\circ\mu(b,d)$ (since this equation holds on the nose for points of $\Omega X$, as can easily be verified by plugging in the definitions of $\mu$ and $\circ$). By Eckmann-Hilton, they must be equal. Explicitly, for any $a,b\in [Y,\Omega X]$ and $1\in[Y,\Omega X]$ the constant map, we have $$\mu(a,b)=\mu(a\circ 1,1\circ b)=\mu(a,1)\circ\mu(1,b)=a\circ b.$$

$\endgroup$
0
$\begingroup$

I too, was trying to answer this question, to understand a talk about homotopy exponents.

First the fact that the two multiplications are homotopic:

The equality $\mu(\gamma_1, \gamma_2) \circ \mu(\Gamma_1,\Gamma_2)=\mu((\gamma_1 \circ \Gamma_1),(\gamma_2 \circ\Gamma_2))$ holds

since both the LHS and RHS applied to $t \in [0,1/2]$ are $\gamma_1(t)\gamma_2(t).$

Both the LHS and RHS applied to $t \in [1/2,1]$ are $\Gamma_1(t) \Gamma_2(t).$

Although this equality is on-the-spot true, the two sided identity is only there for one of the multiplications up to homotopy. So we get that the two multiplications are homotopic.


Here is what we can say about exponents on the h-space and the homotopy groups of the space: if $\circ^k$ is a nullhomotopic map then the kth power map in the fundamental group is 0.

More generally for all the homotopy groups we have that $\cdot: X\times X \to X$ induces $\tilde \mu: \pi_k(X)\times \pi_k(X) \cong \pi_k(X\times X) \to \pi_k(X)$. We have the higher homotopy group composition $\tilde \circ$ taking $a, b: S^n \to X$ to a $\tilde \circ b: S^n \to S^n \wedge S^n \xrightarrow{f \vee g} X$.

The eckman hilton argument is the same: the left hand side and right hand side are the same when applied both below and above the equator of the sphere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.