We’re hiring!
Two permanent positions in pure maths. One in geometry/topology, one open for all areas of pure mathematics. Applications are through mathjobs (link to Job 1, link to Job 2), and are due December 18 (AEDT).
We’re hiring!
Two permanent positions in pure maths. One in geometry/topology, one open for all areas of pure mathematics. Applications are through mathjobs (link to Job 1, link to Job 2), and are due December 18 (AEDT).
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I’ve never been at ease with the Jacobson density theorem, every proof that I’d seen felt somewhat weird and never had any stickiness in my mind. So I came up with my own proof, I hope you like it.
The Theorem (Jacobson Density Theorem}
Let be a field and
a
-algebra (yes unital). Let
be a finite dimensional simple
-module and
(which is a division algebra by Schur’s Lemma). Then the canonical homomorphism from
to
is surjective.
Proof
First, without loss of generality (replacing by a quotient), we can assume that
is faithful.
Let us first deal with the case when is a division algebra. Then
. And
as well, acting by right multiplication. And we are done.
Since is not a division algebra it has a non-zero element
which is not invertible.
If is not nilpotent, then the minimal polynomial
of
factors as
where
. In particular it has two coprime factors, so by the Chinese remainder theorem
has a non-trivial idempotent. In particular,
contains a non-trivial idempotent.
If on the other hand is nilpotent, then we can find
with
,
and
. Since
is irreducible, there exists
with
. Then
and
. So
is not nilpotent, and we can run the argument of the previous paragraph again to conclude that
contains a non-trivial idempotent.
Let be this non-trivial idempotent. We will use the fact that
is an irreducible
-module. This is part of a general fact, that
is a bijection betwen simple
-modules
with
and simple
-modules. Furthermore, from the nature of the constructions of both directions of this bijection, we can conclude that
Now we can perform an induction on dimension. The inductive hypothesis tells us that and
are both surjective. To finish, given the symmetry between
and
, it suffices to show that given any
-hyperplane
with
and any
-line
, there is a nonzero element in
with kernel
and image
. (I thought about this step myself in terms of matrices, but actually writing it up like that feels a little gauche).
The inductive hypothesis applied to allows us to find a nonzero element
with
. Pick nonzero
and
. Since
is simple there exists
with
. Then
does the trick and our proof is complete.
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The following quote is by Matt Emerton (in a comment on MathOverflow)
I think there is a genuine tension between proofs that a professional will like (where professional here may mean professional algebraist!) and ones that are elementary. For professionals, reductions and devissages are easy, natural, and we don’t even think of them as real landmarks in the proof; they are just serve as passages between the key points and ideas. But in writing things out, they can take a lot of words, and seem (as you wrote) mysterious and difficult. I don’t know the best way to deal with this tension.
Interestingly, Matt posted it as part of a discussion about exactly what I wanted to talk about in this post, the teaching of the structure theorem for finitely generated abelian groups, or more generally, of finitely generated modules over a PID.
My personal connection is that I taught this as part of our third-year algebra course this year at the University of Melbourne, and am slated to do so again next year. I think that I did not do a particularly good job of teaching it in 2022, primarily because I got distracted by the reductions and devissages and tried to proceed along those lines as much as possible, when what I have learned is more appropriate for one of these courses is the more prosaic approach involving matrix manipulations. It is with the matrix manipulations (directly proving Smith Normal Form) that I plan to teach this part of the course in 2023 (and beyond, if necessary).
For completeness, allow me to state the professionals’ proof: Split off the quotient by the torsion subgroup to reduce to the torsion case. Then canonically decompose the module into a direct sum of its p-primary components. Then use the fact that is injective over itself to manually split the remaining short exact sequences needed to complete the classification.
While it may not be reasonable to expect a third-year student to follow this proof, I think it is fair to expect any PhD student of mine to be able to understand and execute this proof.
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The Simon Marais Mathematics Competition happened last weekend. It is a maths competition for undergraduate students across Europe, Asia, Africa and Oceania. This post is about problem A4, which I submitted. I’ll talk a bit about where the problem came from, a generalisation, a conjecture and also provide a solution. The entire paper is available on the Marais website, and solutions should be put up there at some time in the near future.
Problem (SMMC 2022 A4)
Let be a positive integer, and let
be an odd integer such that every prime factor of
is larger than
. Prove that
is an integer that has no prime factor in common with .
Origins
Let and let
be the subgroup of monomial matrices (a matrix is a monomial matrix if and only if it has exactly one nonzero entry in each row and column). I show below in my solution that this question is equivalent to the fact that the integer
is coprime to
. Now why would I ever care about that?
This coprimality fact implies that the cohomology of with mod
coefficients is isomorphic to the cohomology of
with mod
coefficients. And I was interested in these cohomology groups because the second cohomology group classifies central extensions, which is what I used to think about back in my PhD days. The group
feels somewhat more “combinatorial” than
, so it is nice to be able to pass information from
to
for free.
Generalisations (known and conjectural)
Let be a split reductive group over
, which I conflate with its
-points below in an abuse of notation. Let
be a maximal split torus and
its normaliser in
. Then
Here is the set of positive roots and the collection of integers
are the exponents of the Weyl group. Then the same argument as in my proof below shows that this fraction is an integer, relatively prime to
.
If we remove the assumption that is split, then I suspect the same conclusion is satisfied, but there is an additional argument needed as the formula for the quotient has additional factors. I have not worked out this argument and really don’t want to resort to case by case arguments, so there is your conjecture (I expect we now need to say
is a maximal torus containing a maximal split torus).
Solution
First we show that the fraction in the question is an integer. Since divides
as a polynomial for all
, the statement only depends on the residue class of
modulo
. Since every prime factor of
is greater than
,
is relatively prime to
. So by Dirichlet’s theorem on primes in arithmetic progressions, we may assume without loss of generality that
is prime.
Let and let
be the subgroup of monomial matrices. Then
By Lagrange’s theorem is an integer. Since
is relatively prime to
, we can further divide by the largest power of
in
and deduce that
is an integer.
Now let be a prime dividing
and let
be a positive integer. To conclude, it suffices to show that the fraction
has zero -adic valuation. Write
, then by the binomial theorem,
Let and
. Since
, we get
We have the inequality , so
For , we therefore get
as from our assumption that
divides
.
Thus in our sum, the term with has a strictly smaller
-adic valuation than every other term, so determines the
-adic valuation of the sum, and we get
completing the proof.
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This is a quick note to prove that two bases of an infinite dimensional vector space have the same cardinality. We freely use the axiom of choice and other standard facts about cardinalities of infinite sets. We will in fact prove the following:
Theorem: Let be a vector space with basis
with
an infinite set. Let
be a linearly independent subset of
. (e.g. a basis of a subspace). Then
.
To prove this, WLOG is a basis of
(by extending
to a basis of
if necessary). For all
, write
Let be the set of pairs
with
. Then
has finite fibres, since the sum above is finite, and
is surjective, since the
lie in the span of the
with
in the image of
, but also the
generate
. Since
is assumed infinite, this is enough to prove that
, as required.
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This post concerns the representation theory of the symmetric group over the complex numbers. Recall that the irreducible representations of the symmetric group are indexed by partitions of
. Let
be the irreducible representation indexed by
. I want to say some words about the theorem that the decomposition of the induced module
is given by the decomposition into eigenspaces under the action of the Jucys-Murphy element.
First, the relevant Jucys-Murphy element is
The way it acts on is not as an element of
but by
This is well-defined since
commutes with
.
What this action defines is a natural transformation from the functor to itself. The induction functor is (bi)-adjoint to the restriction functor and this natural transformation is even simpler to construct on the adjoint side. Recall that if
and
are adjoint functors, then there is an isomorphism
Here refers to the natural transformations from
to itself, and the map in this isomorphism is given by pre- and post-composition by the unit and counit of the adjunction.
And the way that yields a natural transformation from
to itself is very simple, it’s just by its usual action as an element of
. If you transport this natural transformation to a natural transformation of the induction functor via the method I just mentioned, then you get the formula mentioned above.
Now given a pair of adjoint functors and
, a natural transformation
from
to
(and hence from
to
) and a complex number
, we can define a functor
by
and similarly for (this requires some linearity assumptions, but they’re satisfied here. Also you could take generalised eigenspaces if you wanted to, but in our application there is no difference).
When you do this, the functors and
are adjoint:
Proof: Both and
are the
-eigenspace of the action of
on
.
Now apply this to our situation. We also use the following standard fact about the action of the Jucys-Murphy element (as developed e.g. in the Vershik-Okounkov approach):
Consider the decomposition
Then the Jucys-Murphy element acts by the scalar
on
, where
is the content of the box
added to
to get
.
Now translating this statement via the above yoga onto the adjoint side, we get
In the decomposition
the Jucys-Murphy element acts by the scalar
on
, where
is the content of the box
added to
to get
.
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Martina Lanini and I recently posted our preprint Singularities of Schubert varieties within a right cell to the arXiv. In it, we show that every singularity which appears in a type A Schubert variety appears between two permutations lying in the same right cell. This shows that any behaviour controlled by the singularities of Schubert varieties manifests itself within a Specht module. Some exmples are discussed.
The work was conducted during our recent visit to the thematic trimester program on representation theory at the Institut Henri Poincaré in Paris. I spent an enjoyable first month there before returning to Australia. Originally I was scheduled to be on a plane right now to return to Paris for the end of the program, but alas this is no longer possible. Oh well.
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The University of Melbourne maths department is hiring. All our ads can be found on mathjobs.
In pure maths, we have a continuing position in analysis. Applications should be submitted by the earlier deadline of 25 October to ensure full consideration.
There are also other positions available in applied and statistics, as well as a temporary position open to all areas.
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There is a new (established in the last couple of years) mathematical institute based at the University of Sydney. I want to call attention to a visitor program they run, for mathematical visits to Australia.
They fund research visits of at least one month’s duration. At least two weeks must be spent at the University of Sydney, but the rest can be taken anywhere in Australia.
Funding rounds for these research visits happen periodically. Currently there is one open for visits within the period April-December 2020, with applications closing on 28 July.
Anyone interested should check out the research intstitute’s webpage.
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