# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 27th 2018

am starting something. Just saving for the moment, not done yet

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeSep 27th 2018
• (edited Sep 27th 2018)

okay, I have recorded the basic definition/statement

I tried to bring out the nice picture as in tom Dieck 09, 6.2, in particular I focused, for the moment, on field extensions from $\mathbb{Q}$.

I have taken the liberty of calling the direct sum of a $\widehat k$-irrep with all its distinct Galois translates its Galois group averaging (here) since I find that saying this helps to see at a glance what’s really going on here.

I have also taken the liberty of giving an alternative characterization of this “Galois group averaged representation”, namely as the smallest rep that is in the kernel of $\Psi^n - id$ for $n$ coprime to ${\vert G\vert}$. Because this makes the relation to the Adams conjecture manifest. Hope I got this right.

But representation theorists please feel invited to criticize.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeSep 27th 2018
• (edited Sep 27th 2018)

I have also added a remark (here) that highlights the close similarity of the construction to the J-homomorphism and the Adams conjecture.

So I highlighted in particular the (open?) question (raised in another thread here) of whether indeed the Schur index construction is the incarnation of the equivariant J-homomorphism over the point, and if the “Galois group averaging” involved is the equivariant Adams conjecture-statement on the point – at least for those groups $G$ for which $A(G) \overset{\beta}{\to} R_{\mathbb{Q}}(G)$ is surjective – because it looks directly analogous.

This may be straightforward to check by chasing through Segal’s proof of $\mathbb{S}^0_G(\ast) \simeq A(G)$. If this question turns out to be really open, I’ll sit down and do that. But I’d much rather just cite it from somewhere. If it is indeed true.