2))$ has odd dimension?

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Feb 1, 2011 at 1:25	comment	added	Greg Friedman		Martin, if you contact me with your full name, I'll make sure to give you proper credit for this argument in the paper I'm working on. (Otherwise I'll have to credit you as "Martin O" at mathoverflow :-) )
Jan 30, 2011 at 6:35	comment	added	Greg Friedman		Very nice argument. Thank you. Hmmm. I wonder if it can be adopted to intersection homology using Goresky's intersection cohomology operations. I'm not sure if the Bockstein's would work out, though, because of the intricacies of coefficients in intersection homology (in particular, $IC(X;G)$ is not necessarily the same as $IC(X)\otimes G$)
Jan 30, 2011 at 6:30	vote	accept	Greg Friedman
Jan 29, 2011 at 0:14	comment	added	Martin O		@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.
Jan 29, 2011 at 0:08	comment	added	Martin O		@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 $e(\xi)=\tilde{\beta}w_{2k}(\xi)$.)
Jan 28, 2011 at 22:24	comment	added	John Klein		@Martin: can you explain why the kernel is the radical?
Jan 28, 2011 at 16:10	comment	added	John Klein		The fact that it's not possible leads to an interesting corollary: given an oriented vector bundle $\xi$ of rank $2k+1$ over a connected closed manifold of dimension $2k+1$, then the top Stiefel-Whitney class $w_{2k+1}(\xi)$ is trivial. The proof: the zero section map $M \to M^{\xi}$ (the target is the Thom space) represents the Euler class. It also coincides with Jeff's map $H_{2k+1}(D(\xi)) \to H_{2k+1}(D(\xi),S(\xi))$. QED. Is this a known result? I couldn't find it in the literature.
Jan 28, 2011 at 0:57	history	answered	Martin O	CC BY-SA 2.5