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Stream: deprecated: mathematics

Topic: algebraic K-theory of the integers

John Baez (Dec 09 2020 at 01:34):

I've been working hard trying to finish two papers before the end of the year. But I have to goof off a little or I'll go crazy. So I wrote a quick intro to this:

• The algebraic K-theory of the integers.

John Baez (Dec 09 2020 at 01:47):

Basic idea:

The category of groups $\mathbb{Z}^n$ and isomorphisms between these is symmetric monoidal under $\oplus$. You can build a space out of simplexes where the 0-simplexes are objects of this category, the 1-simplexes are morphisms, the 2-simplexes are commutative triangles, the 3-simplexes are commutative tetrahedra, and so on forever. This space has an operation, coming from $\oplus$, that obeys the commutative monoid axioms up to homotopy. If you 'group complete' this space by throwing in formal inverses, you get a space that's an abelian group up to homotopy. It's called the algebraic $K$-theory spectrum of the integers.

Ian Coley (Dec 09 2020 at 06:35):

I see a lot of questions -- how rhetorical are they?

Morgan Rogers (he/him) (Dec 09 2020 at 11:03):

John Baez said:

I've been working hard trying to finish two papers before the end of the year. But I have to goof off a little or I'll go crazy.

This is exceptionally relatable to me right now :joy:

John Baez (Dec 09 2020 at 19:50):

Ian Coley said:

I see a lot of questions -- how rhetorical are they?

There are 3 questions. The first two I'd like answers to:

Is this right? Are these the right homomorphisms?

The third is a puzzle that apparently everyone is too shy to publicly answer on the blog:

Notice anything interesting about these numbers less than and relatively prime to 24?

which is sad because it denies me the chance to say something interesting.

Morgan Rogers (he/him) (Dec 09 2020 at 21:36):

Are you looking for "they're almost all prime" for that last question? :joy:

John Baez (Dec 09 2020 at 21:37):

Yes, and that would start another conversation.

John Baez (Dec 09 2020 at 21:37):

(On the blog.)

John Baez (Dec 09 2020 at 21:37):

This one of the famous properties of the numbers 24... and 30.

Morgan Rogers (he/him) (Dec 09 2020 at 21:42):

Alright, as you wish..!

John Baez (Dec 09 2020 at 22:27):

Okay, you gave me the excuse to talk about some weird stuff.

John Baez (Dec 10 2020 at 07:09):

I'm trying to learn some stuff about the algebraic K-theory groups $\mathrm{K}_n(\mathbb{Z})$ and stable homotopy groups $\pi_n^s$.

John Baez (Dec 10 2020 at 07:09):

I'm interested in the case where $n = 4m-1$, since this is when Bernoulli numbers show up, and some other fun stuff.

John Baez (Dec 10 2020 at 07:11):

There's a certain chunk of $\pi_n^s$ called $\mathrm{Im}(J)_n$, the image of the J-homomorphism, and Adams showed that when $n = 4m-1$ it's a cyclic group whose order is connected to Bernoulli numbers.

John Baez (Dec 10 2020 at 07:16):

The Wikipedia link states the facts pretty clearly. Using my notation, the order of $\mathrm{Im}(J)_{4m-1}$ is the denominator of

John Baez (Dec 10 2020 at 07:16):

$B_{2m}/4m$

John Baez (Dec 10 2020 at 07:17):

where $B_{2m}$ is a Bernoulli number.

John Baez (Dec 10 2020 at 07:17):

Here are some examples:

John Baez (Dec 10 2020 at 07:18):

$m = 1$, $B_{2m} = 1/6$, so the denominator of $B_{2m}/4m$ is $4 \times 6 = 24$, so $\mathbb{Z}/24 \cong \mathrm{Im}(J)_3 \subseteq \pi_3^s$. In this case $\mathbb{Z}/24$ is all of the stable homotopy group $\pi_3^s$.

John Baez (Dec 10 2020 at 07:23):

$m = 2$, $B_{2m} = -1/30$, so the denominator of $B_{2m}/4m$ is $8 \times 30 = 240$, so $\mathbb{Z}/240 \cong \mathrm{Im}(J)_7 \subseteq \pi_7^s$. In this case $\mathbb{Z}/240$ is all of the stable homotopy group $\pi_7^s$.

John Baez (Dec 10 2020 at 07:26):

$m = 3$, $B_{2m} = 1/42$, so the denominator of $B_{2m}/4m$ is $12 \times 42 = 504$, so $\mathbb{Z}/504 \cong \mathrm{Im}(J)_{11} \subseteq \pi_{11}^s$. In this case $\mathbb{Z}/504$ is all of the stable homotopy group $\pi_{11}^s$.

John Baez (Dec 10 2020 at 07:28):

I have no idea why $\mathrm{Im}(J)$ works this way, and that's what I'm trying to understand.

John Baez (Dec 10 2020 at 07:29):

But anyway, it looks like Bernoulli numbers might be a bit of a red herring, since all we care about is the denominators of Bernoulli numbers.

John Baez (Dec 10 2020 at 07:30):

Adams proved these were connected to $\mathrm{Im}(J)$ using Von Staudt's theorem.

John Baez (Dec 10 2020 at 07:31):

This theorem says the denominator of $B_{2m}$ is the product of all primes $p$ such that $p-1$ divides $2m$.

John Baez (Dec 10 2020 at 07:31):

Let's test it out.

John Baez (Dec 10 2020 at 07:34):

$m = 1$. The primes $p$ such that $p-1$ divides $2m = 2$ are 2 and 3, so the denominator is $2 \times 3 = 6$.

John Baez (Dec 10 2020 at 07:35):

$m = 2$. The primes $p$ such that $p - 1$ divides $2m = 4$ are 2, 3 and 5 so the denominator is $2 \times 3 \times 5 = 30$.

John Baez (Dec 10 2020 at 07:41):

$m = 3$. The primes $p$ such that $p -1$ divides $2m = 6$ are 2, 3 and 7 so the denominator is $2 \times 3 \times 7 = 42$.

John Baez (Dec 10 2020 at 07:46):

Yup, it's working. So maybe Bernoulli numbers are less important than these products of primes.

John Baez (Dec 10 2020 at 07:47):

Note that the order of $\mathrm{Im}(J)_n$ winds up being a power of two times a bunch of distinct odd primes.