This is a question about signs.


  • a based space $(X,x_0)$,
  • a topological group $G$
  • acting on $X$ from the left, so that the basepoint $x_0$ is fixed,
  • a based map $\alpha\colon S^p\to G$ ($p\geq1$), and
  • a based map $\beta\colon S^q\to X$ ($q\geq 1$).

I define (homotopy classes of) based maps $\gamma,\gamma'\colon S^{p+q}\to X$ as follows.

  1. We have a fiber sequence $$ X\xrightarrow{i} X\times_G EG \to BG$$ with section $s\colon BG\to X\times_G EG$ determined by the basepoint $x_0$. Let $\gamma$ be the unique homotopy class such that $$ i\circ \gamma:=[s\circ \widetilde\alpha, i\circ \beta],$$ (Whitehead product) where $\widetilde\alpha\colon S^{p+1}\to BG$ is the adjoint to $\alpha$.

  2. Consider the composite $$ S^p\times S^q \xrightarrow{\alpha\times\beta} G\times X\to X,$$ using the action $G\curvearrowright X$. Since $G$ fixes $x_0$, this composite factors through a map $$\kappa\colon H:=(S^p\times S^q)/(S^p\times *)\to X.$$ We define $$\gamma':= \kappa\circ \sigma,$$ where $\sigma\colon S^{p+q}\to H$ is a section-up-to-homotopy of the quotient map $H\to S^p\wedge S^q=S^{p+q}$. (It looks like $\sigma$ involves a choice, but in fact we can fix a "canonical" choice explicitly: it corresponds to the "obvious null-homotopy" of $[0,\iota_q]$. When $q\geq2$, we can fix $\sigma$ by requiring that the composite with the projection $H\to S^q$ be null.)

I can convince myself that $\gamma'\sim \gamma$. I can also convince myself that $\gamma'\sim -\gamma$. So which is it? Does the sign depend on $p$ or $q$?

Probably, you can't answer that without knowing about all the conventions I have used, that I haven't bothered to mention (or am completely oblivious to). The real question is: is there a reference that deals with this kind of formula, and is likely to be reliable about signs, and explicit about choices of convention which may affect signs? Alternately, do you know some example which clarifies how the signs should go?

  • 1
    $\begingroup$ If we think of $G$ as acting on $\Omega X$ then it looks like there's a map $G \star \Omega X \longrightarrow X$ which also induces a pairing on homotopy groups like the one you described. This seems related to both of your constructions since (1) the Whitehead product comes from the canonical pairing $\Omega X \star \Omega X \rightarrow X$ for any space, and (2) there should be some kind of naturality relating this construction to fiber sequences, e.g. path-loops and homotopy orbits. Not sure if this is useful... $\endgroup$ – Dylan Wilson Feb 21 '16 at 23:06
  • $\begingroup$ @DylanWilson I agree with your line of thinking. The point, I guess, is that the only real sign issues are lurking in the "adjoint" relation $S^p\to G$ vs. $S^{p+1}\to BG$, which can be fixed relating the path fibration $P(BG)\to BG$ to the fibration $EG\to BG$. ... $\endgroup$ – Charles Rezk Feb 23 '16 at 1:04
  • $\begingroup$ Of course, this means I have to figure out things like the convention for path composition (i.e., product on $\Omega X$): is $\gamma*\delta$ in "temporal order" (put $\gamma$ on $[0,1/2]$) or "composition order" (put $\gamma$ on $[1/2,1]$, cause we are thinking of it as a morphism $\gamma(0)\to \gamma(1)$, so the output of $\delta$ is the input of $\gamma$)? $\endgroup$ – Charles Rezk Feb 23 '16 at 1:06
  • $\begingroup$ Also, what is a good reference for the Whitehead product? I thought G. Whitehead's book would be good, but I can't figure out where his basepoint is! (I realize it is not the same Whitehead.) $\endgroup$ – Charles Rezk Feb 23 '16 at 1:07
  • $\begingroup$ I'm still mystified but maybe the following references will be useful: (1) G Whiteheads "generalization of the hopf invt" section 3, (2) caftan 1958/59 expose 5 section 2 (especially the statement of a lemma due to Samelson identifying the sign that appears when you use the adjoint isomorphism), (4) the only relatively recent references I found are Baues algebraic homotopy sec 15 and B. Gray's paper 'on generalized whitehead products' $\endgroup$ – Dylan Wilson Feb 23 '16 at 4:15

Your Answer

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

Browse other questions tagged or ask your own question.