User tamas hausel - MathOverflow

Answer by Tamas Hausel for Is there an algebraic construction of the Quillen (determinant) Line Bundle?

2011-06-20T04:48:33Z

For $G={\rm PGL}_n({\mathbb C})$ there is no complex algebraic Quillen line bundle on the complex character variety $Y$ as the cohomology class $[\omega]=\alpha_2\in H^2(Y;{\mathbb Q})$ is not pure by Proposition 4.1.8 in http://arxiv.org/abs/math.AG/0612668.

EDIT: It is instructive to think about the case $G={ GL}_1({\mathbb C})$. Then the character variety is $M_B=GL_1^{2g}$ the complex torus while $M_{DR}$ is an affine bundle over $Jac(C)$. Analytically $M_B\cong M_{DR}$ but the universal bundle $L$ on $M_{DR}\times C$ which is just the pull back of the Poincare bundle from $Jac(C)\times C$ is not algebraic on $M_B$, for example because the cohomology class $[\omega]$ of the symplectic form (which shows up in $c_1(L)^2$) has weight $2$ in $H^2(M_{DR};{\mathbb Q})$ but has weight $4$ (homogeneous weight $2$) in $H^2(M_B;{\mathbb Q})$. Thus in particular there is no complex algebraic Quillen bundle on $M_B$ with first Chern class $[\omega]$.

EDIT 2: To take David's example let $g=1$, then $M_B={\mathbb C}^* \times {\mathbb C}^*$ which does not have non-trivial pure cohomology in $H^2(M_B;{\mathbb Q})$ and even the Picard group is trivial (see an argument here); thus $M_B$ does not support any non-trivial algebraic line bundle.

Answer by Tamas Hausel for Homotopy type of Hilbert schemes of points of $\mathbb C^2$

2011-04-21T21:53:04Z

It is a general folklore result, that if ${\mathbb C}^\times$ acts on a smooth complex variety $X$ so that the fixed point set $X^{{\mathbb C}^\times}$ is proper and the limit ${{\rm lim}_{\lambda \to 0} \lambda z } $ exists for every $z\in X$, then the downward flow $$D:=\{ z\in X | {\rm lim}_{\lambda \to \infty} \lambda z \mbox{ exists}{\}}$$ is a retract of $X$. This can be proved by Morse theory arguments as in Kirwan's book or you can prove it by first showing that the imbedding induces

$$H_*(D;{\mathbb Z})\to H_{*}(X;{\mathbb Z})$$ an isomorphism (by induction with respect to an appropriate ordering of the set of components of $X^{{\mathbb C}^\times}$). Similarly you can show that the fundamental groups are also isomorphic. Then the relative Hurewitz theorem will tell you that they are weakly homotopy equivalent, and as they are varieties so CW-complexes, Whitehead's theorem imply that they are homotopy equivalent.

For the Hilbert scheme you can use the ${\mathbb C}^\times$ action induced from the dilation on ${\mathbb C}^2$ mentioned above, which will have the required property, due to the fact that the Hilbert-Chow morphism is proper.

Answer by Tamas Hausel for Decomposing tensor products of irreducible representations of reductive groups over a finite field

2010-01-20T07:15:52Z

Theorem 1.4.1 in arxiv:0810.2076 answers some of your questions for generic semisimple irreducible representations. Emmanuel Letellier has hitherto unpublished results where he does answer your question for all generic irreducible representations in terms of intersection cohomology of certain quiver varieties. We did not know about other results on the representation ring of $GL_n({\mathbb F}_q)$. EDIT: but see Victor's answer for related results of Lusztig. EDIT 2 (added 16/03/11) Letellier's paper is now available at: http://arxiv.org/abs/1103.2759

Answer by Tamas Hausel for homology of a symplectic leaf in GL(n)

2010-10-21T10:18:33Z

Below now is a proof that for sufficiently generic $z$, $b_1(Y(z))=n-1$.

First we show $b_1(Y(z))\geq n-1$.

Let $d=(d_1,d_2..,d_{n-1}): Y(z)\to({\mathbb C}^*)^{n-1}$. One can explicitly lift the loops in $H_1(({\mathbb C}^*)^{n-1})$ as follows. Fix $i$ between $1$ and $n-1$. Let $e(t)=exp(2\pi \sqrt{-1} t)$ and $$B(t):=\left(\begin{array}{cc} e(t) & 1\\ e(t)(z_i+z_{i+1}-e(t))-z_iz_{i+1} & z_i+z_{i+1}-e(t) \end{array}\right).$$ Then the characteristic polynomial of $B(t)$ is $(\lambda-z_i)(\lambda-z_{i+1})$. Form an $n\times n$ matrix $A(t)$ as having $z_1,\dots,z_{i-1}$ on the main diagonal, then $B(t)$, then $z_{i+2},\dots,z_{n}$ and zero everywhere else. This matrix has the property that $A(t)\in Y(z)$ (as we assumed generic z) and $d_j(A(t))=z_1\cdot\dots\cdot z_j$ unless $j=i$ when $d_i(A(t))=e(t)z_1\cdot\dots\cdot z_{i-1}$. Thus $A(t)$ is a loop in $Y(z)$ covering the $i$th generator of $H_1(({\mathbb C}^*)^{n-1})$.

Thus we find that $d_*:H_1(Y(z))\to H_1(({\mathbb C}^*)^{n-1})$ is surjective so $b_1(Y(z))\geq n-1$.

To argue that $b_1\leq n-1$ we first show that the divisors $D_i:=d_i^{-1}(0)$ are irreducible on $X(z)$ for sufficiently generic $z$. We note that $D_i$ is irreducible in $M_{nxn}$ which can be easily seen over a finite field by counting points on $D_i$. This implies that for a sufficiently generic $z$ the divisor $D_i$ is irreducible on $X(z)$. When $z_i\neq z_j$ for $i\neq j$ we see that $X(z)$ fibers over the full flag variety with contractible fibers, so it is simply connected. From the homology Gysin sequence we get that removing an irreducible divisor from a smooth variety can increase $b_1$ by at most one. By induction we can conclude that $b_1(Y(z))\leq n-1$.

As an example take $n=2$. Here we are removing a copy of ${\mathbb C}^*$ from $X(z)$ an affine line bundle over ${\mathbb P}^1$. This will have Betti numbers $b_1=1$ and $b_2=2$ with weights $q,q,q^2$ respectively, to give Serre-polynomial $q^2+1$. We will have the map $d=d_1:Y(z) \to {\mathbb C}^{*}$ . The fibers of this map will generically be a ${\mathbb C}^{*}$ and will have two singular fibers, which will give the non-trivial fundamental group. The point is that one can lift the loop generating $\pi_1({\mathbb C}^*)$ as generically $d_1$ is a fibration with connected fibers.

Answer by Tamas Hausel for Finitely many arithmetic progressions

2010-05-20T10:21:47Z

This was one of my favourite problems in high school. My proof went like this: if you look at the problem modulo n where n is the least common multiple of the differences of the arithmetic progressions then you can rephrase the problem as follows: if the vertices of a regular n-gon are partitioned into regular k-gons centered at the origin then two of them will have the same size. To prove this arrange the regular n-gon to have vertices at the nth roots of unity in the complex plane and assign the monic polynomial to every individual k-gon whose roots are exactly the vertices of the polygon. This way you will get the expression $x^n-1=(x^{k_1}-\zeta_1)(x^{k_2}-\zeta_2)\cdots$. Multiplying out the RHS we see that if $k_1$ is the least of the $k_i$'s then the only way to cancel the term of $x^{k_1}$ from the RHS is to have another $k_i=k_1$ proving the claim.

Answer by Tamas Hausel for Is the wedge product of two harmonic forms harmonic?

2009-11-07T15:26:55Z

Interestingly in (24) of hep-th/9603176 it is mistakenly claimed that the wedge product of harmonic forms is automatically harmonic. Because it is false we still do not know the predicted existence of those middle dimensional $L^2$ harmonic forms on these non-compact complete hyperkahler manifolds.

Comment by Tamas Hausel

2011-06-20T13:59:24Z

David, the universal bundle is only algebraic in the $M_{DR}$ algebraic structure - it is not algebraic in the character variety algebraic strucure, as its Chern classes are not pure.

Comment by Tamas Hausel

2010-09-29T16:24:46Z

The chessboard problem can be nicely continued by proving that the chessboard can be tiled with dominos if two differently coloured squares are removed. Check out the solution of puzzle B at gurmeetsingh.wordpress.com/2008/09/12/…

Comment by Tamas Hausel

2010-01-20T16:31:12Z

Interesting. Any chance to find the result for the character ring of $PGL_2({\mathbb F}_q)$ somewhere explicitly?

Comment by Tamas Hausel

2009-11-07T18:39:42Z

No, the individual 2-forms are harmonic, and on Taub-Nut is is in $L^2$ as well. One can show that there are no other ones as in arxiv.org/abs/math/9909002 and arxiv.org/abs/math/0207169