User martin o - MathOverflow

Answer by Martin O for How to caculate the internal hom of supermanifolds?

2011-02-18T18:10:09Z

(Sorry for not answering this on the previous post, you asked this question before. By the way, you didn't say which paper you are reading...)

$Map(\mathbb R^{0|0},M)=M$. Trivial.

$Map(\mathbb R^{0|1},M)=\Pi T M$. ($TM$ is the tangent bundle of $M$, and $\Pi$ reverses grading of a vector bundle. So $\Pi T M$ denotes the total space of the degree-reversed tangent bundle of $M$. If $\dim M=d|\delta$, then $\dim \Pi T M=d+\delta|d+\delta$.)

Since

$Hom(S, Map(\mathbb R^{0|n},M)) = Hom(S\times \mathbb R^{0|n},M)$ $= Hom(S\times \mathbb R^{0|n-1}\times \mathbb R^{0|1},M) = Hom(S\times \mathbb R^{0|n-1},\Pi T M)$,

we obtain inductively $Map(\mathbb R^{0|n},M)=(\Pi T)^n M$. However, the interesting thing about $Map(\mathbb R^{0|n},M)$ is that it has an action by $Diff(\mathbb R^{0|n})$, the supermanifold of invertible maps from $\mathbb R^{0|n}$ to itself. I guess this is not so visible if you write $(\Pi T)^n M$, since this was obtained by destroying the symmetry in the odd coordinates.

For a description which keeps the symmetry: $Map(\mathbb R^{0|2},M)$ is the pullback of $\Pi( T M\oplus T M) \to M$ along $TM\to M$, which I learned from Dan Berwick Evans at Berkeley. I would guess this is as explicit as it gets in general, and that probably more difficult pullbacks squares involving $\Pi T M$ exist for $Map(\mathbb R^{0|n},M)$ with bigger $n$.

One can find a description and discussion of $Map(\mathbb R^{0|n},M)$ in the paper with the nice title "Differential gorms, differential worms", Denis Kochan, Pavol Severa arXiv:math/0307303.

Answer by Martin O for Morphisms between supermanifolds R^{0|1}→R^{0|1}

2011-02-09T23:47:12Z

You are right that the set of supermanifold morphisms $Hom(\mathbb R^{0|1},\mathbb R^{0|1})$ to itself is $\mathbb R^1$. However, one can define for supermanifolds $X,Y$ with $\dim X=0|d$ a supermanifold $map(X,Y)$ of morphisms from $X$ to $Y$, by $Hom(Z,map(X,Y))=Hom(Z\times X,Y)$ for all supermanifolds $Z$.

And $map(\mathbb R^{0|1},\mathbb R^{0|1})=\mathbb R^{1|1}$.

Answer by Martin O for Are there oriented $4k+2$ manifolds such that $im(H_{2k+1}(M; Z/2)\to H_{2k+1}(M, \partial M; Z/2))$ has odd dimension?

2011-01-28T00:57:04Z

I claim it is not possible. The image is the rank of $H_{2k+1}(M;\mathbb Z_2)/rad$, where $rad$ is the radical of the intersection form on $H_{2k+1}(M;\mathbb Z_2)$.

The intersection form on $H_{2k+1}(M;\mathbb Z_2)/rad$ is hyperbolic, i.e. has a "symplectic" basis, therefore this vector space has even dimension.

Let me try to prove that it is hyperbolic: the tricky point is to show that all classes square to zero, i.e. $\langle x^2,[M,\partial M]\rangle =0$ for all $x\in H^{2k+1}(M,\partial M;\mathbb Z_2) $.

Now $\langle x^2,[M,\partial M]\rangle =\langle \beta Sq^{2k}x,[M,\partial M]\rangle=\langle Sq^{2k}x,\beta [M,\partial M]\rangle=0$ where $\beta$ denotes the cohomology respectively homology Bockstein.

Answer by Martin O for Equivariant Surgery problem

2010-07-27T19:50:55Z

Yes. You need to extend the map to $BG$ over the surgery cobordism, which is possible.

First let me add that all your $G$-actions seem to be free and that the boundary of $V$ would consist of $r$ copies of $M$.

Now for the surgeries: first take connected sum of V/G with 2 copies of $S^1\times S^2$'s (surgeries on embeddings $S^0\times D^3$) and extend the map to BG as $S^1\times S^2\to S^1$ and then to generators of $G=\pi_1(BG)$ to make $V'/G\to BG$ 1-connected. Then find $S^1$'s in $V'/G$ generating the kernel of $\pi_1$, thicken to embeddings of $S^1\times D^2$ (they have trivial normal bundle) and do surgery on them, extending the map to $BG$ by nullhomotopies of their images in $BG$.

Answer by Martin O for Collapsing contractible subsets of the two-disk.

2010-07-13T20:24:06Z

The equivalence classes should be closed subsets of the disk to make the quotient Hausdorff.

Answer by Martin O for Thematic Programs for 2010-2011?

2009-10-24T11:54:02Z

The Hausdorff Research Institute for Mathematics, Bonn, Germany has:

Future Trimester Programs

* Geometry and dynamics of Teichmüller space
  May - August 2010
* On the Interaction of Representation Theory with Geometry and Combinatorics
  January - April 2011
* Analysis, and Numerics for High Dimensional Problems
  May - August 2011

Future Junior Programs

* Algebra and Number Theory
  January - April 2010
* Stochastics
  September - December 2010

Here's a link to their webpage.

Comment by Martin O

2011-02-22T10:19:07Z

I think your second explanation is good. Also, it is maybe even better to consider the action of the whole supersemigroup $Map(\mathbb R^{0∣n},\mathbb R^{0∣n})$ instead of the action of the subgroup $Diff(\mathbb R^{0∣n})$ of invertible maps, which can be defined using functor of points again. It is true that associativity follows from associativity of composition of maps between supermanifolds via some abstract nonsense.

Comment by Martin O

2011-02-21T11:27:36Z

Updated my answer!

Comment by Martin O

2011-02-21T08:09:20Z

One reference could be Brylinski's book (and he gives older references): Loop Spaces, Characteristic Classes and Geometric Quantization, Section 2.1

Comment by Martin O

2011-02-10T08:30:09Z

Thanks for spotting the typo, which I corrected.

Comment by Martin O

2011-02-07T00:17:27Z

@Mark: One can consider the pair $(K(\mathbb Z,1),X)$ as in the exercise, and then apply Theorem 4.37 and the long exact sequences in homology and homotopy.

Comment by Martin O

2011-02-06T20:47:35Z

Hatcher has an exercise leading to a different proof: p.390, ex. 23.

Comment by Martin O

2011-01-29T00:14:02Z

@John's second comment: the adjoint of the intersection form $H_{2k+1}(M;\mathbb Z_2) \times H_{2k+1}(M;\mathbb Z_2)\to\mathbb Z_2$ is given by $H_{2k+1}(M;\mathbb Z_2) \to H_{2k+1}(M,\partial M;\mathbb Z_2)\cong H^{2k+1}(M;\mathbb Z_2)$ where we used Poincaré duality. The kernel of the map is the radical of the intersection form, i.e. all classes which have zero intersection with everything.

Comment by Martin O

2011-01-29T00:08:06Z

@John's first comment: If the bundle is oriented, M needs to be oriented as well. The Euler class of an odd-dimensional bundle is known to be 2-torsion, so even the Euler class is zero here. (Hatcher has an exercise in the vector bundle book that for every oriented $(2k+1)$-dimensional bundle $\xi$ one has $e(\xi)=\tilde{\beta}w_{2k}(\xi)$.)

Comment by Martin O

2011-01-26T21:33:26Z

You need to start by a map from $N\to \mathbb R P^\infty$ classifying the double cover $M\to N$. This factors through $\mathbb R P^j$ with $j = 2\dim M + 1$, can be perturbed to an embedding, and if $M$ is a non-trivial-cover, then the map $N\to \mathbb R P^j$ is non-trivial on fundamental groups, and so the image cannot be contained in a ball.

Comment by Martin O

2010-12-15T21:49:10Z

Susan Tolman's paper arxiv.org/abs/0903.4918 also considers the 6-dimensional case. From the argument (Remark 2.11) with the $\chi_y$-characteristic one should be able to deduce the other Chern classes of your putative example.

Comment by Martin O

2010-11-15T23:38:07Z

It's a kernel of a surjective group homomorphism to $Z_2$.

Comment by Martin O

2010-09-23T01:21:00Z

@Peter: According to the online etymology dictionary, the two meanings of pole come from two different latin words: palus respectively polus (similar as in Italian). So my guess is that at the time of invention of the term the words were different. Still you might be right that originally all these words have its origin in the same greek word.

Comment by Martin O

2010-09-21T23:21:26Z

This is a nice explanation, but I don't think that this is the correct one since in French (and German) the translation of your use of pole would not be "pôle" (or "Pol"). I guess it is rather related to the poles of a magnet.