Subject: Re: 2 on htpy class names
Date: Mon, 7 Oct 2002 20:50:54 -0500
From: Bill Richter
To: dmd1@lehigh.edu
From: John Rognes
Subject: eta, nu, sigma.
If alpha, beta, gamma and delta are reserved for other, generic
purposes, such as writing about Toda brackets, epsilon is the first
letter free for use.
I like this non-ImJ Greek alphabetical order idea, John! But:
Then if mu, zeta (and possibly rho) are somehow already referring
to elements in the image of J,
Mark Mahowald tells me that zeta (generator of 11 stem Z/8) is in ImJ.
But mu (9 stem) isn't in ImJ, and neither is nu-bar (8 stem). They
can't be, because 8 & 9 are Z/2 degrees, and sigma eta & sigma eta^2
are clearly in ImJ by naturality. Dunno about rho yet.
iota to the identity, and eta, nu and sigma to the Hopf invariant
one elements, the next spare letters in the Greek alphabet are
theta, kappa, lambda and xi. If theta was occupied for odd primary
purposes, this makes kappa the next available letter.
Actually Toda uses theta for an unstable element in the 12 stem, so
kappa in the 14 stem is fine for your Greek alphabetical order idea.
Lambda is next, and I don't think Toda uses that for an element, but
we could say that lambda is a generic Greek letter like alpha, beta,
gamma, delta, and iota. There's the Lambda algebra, and lambda
operations in K theory, and Boardman-Steer's Hopf invariants...
However, xi comes before rho, and xi = {sigma_12, nu, sigma} is in the
18 stem, it's h_2 h_4 in fact.
Could Toda be alphabetical by "order of difficulty", and not by stem?
xi is a pretty simple Toda bracket (like Barratt's original bracket
nu' = {eta_3, 2iota, eta}): xi might've occured to Toda early on.
But kappa is much harder than mu = {eta_3, 2iota, sigma'''}; kappa is
Toda's first a 4-fold Toda bracket.
However, Mahowald tells me that mu is "almost" in ImJ:
First mu and eta mu are not strictly image of J elements. Under the
map S^0 -> bo it is elementary to see that mu and eta mu map
nontrivial.
So by extending your ImJ convention a little, mu deserves your ImJ
exception, and can be excluded from the Greek alphabetical ordering.
And for the first twenty-odd two-primary stable stems, these are
all the letters Toda needs.
Yeah, 19 two-primary stems (unstable as well as stable) in his book,
although I think Toda & Mimura made it up to the 30 stem or so.
Bill