most recent 30 from http://mathoverflow.net 2013-05-18T20:16:25Z http://mathoverflow.net/feeds/question/27985 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27985/splitting-of-the-universal-coefficients-sequence Jeff Strom 2010-06-13T01:26:46Z 2010-06-19T13:20:11Z

<p>The really beautiful way to prove the Universal Coefficients theorem, to my taste, is to use the fibration sequence $K(\mathbb{Z}, n) \to K(\mathbb{Z}, n) \to K(\mathbb{Z}/k, n)$ (I'm using $\mathbb{Z}/k$ coefficients for simplicity) to give a long exact sequence of cohomology, and split it into short exact sequences. (I'm pretty sure I learned this from Adams' Generalized Homology...).</p> <p>Anyway, this gives the desired exact sequence, but I don't know how to get the (nonnatural) splitting that the homological algebra derivation provides.</p>

http://mathoverflow.net/questions/27985/splitting-of-the-universal-coefficients-sequence/27996#27996 Torsten Ekedahl 2010-06-13T05:24:46Z 2010-06-13T07:08:26Z

<p>I would claim that the splitting (and indeed the whole universal coefficient theorem) is not really a topological theorem. If we take the homological version one really works with the chain complex $C_\ast(X)$ in the derived category of $\mathbb Z$-complexes. We then have $C_\ast(X,M)=C_\ast(X)\bigotimes M$ but as $C_\ast(X)$ is free this equals the derived tensor product $C_*(X)\bigotimes^{\mathbb L} M$ and hence is a formula in the derived category. One can then use the fact that in the derived category of $\mathbb Z$-modules every complex is isomorphic to the sum of its (shifted) homology: $C\cong \bigoplus_nH_n(C)[n]$ so that $$ C_*(X)\bigotimes^{\mathbb L} M \cong \bigoplus_n(H_n(X)\bigotimes^{\mathbb L} M)[n] $$ and as $A\bigotimes^{\mathbb L} M\cong A\bigotimes M\bigoplus \mathrm{Tor}^1(A,M)[1]$ we get the universal coefficient formula including the splitting.</p> <p>This idea also demonstrates why the splitting is not canonical. We may for instance consider a group $G$ acting on $X$. We then get at complex $C_*(X)$ in the derived category of $G$-modules and a complex in that category is in general not isomorphic to the sum of its homology.</p> <p>On the other hand, this technique can be used (essentially) each time some invariant of a topological space $X$ only depends on its chain complex in a way that takes quasi-isomorphisms to isomorphisms. The conclusion is that it only depends on the homology of $X$. A nice example is the homology of the $n$'th symmetric product of $X$. It turns out to be the homology of a complex constructed functorially from $C_*(X)$ and exactly in a way that preserves quasi-isomorphisms. Hence it only depends on the homology of $X$ (and one can also give explicit formulas).</p> <p>However, the method that you declare a fondness for is also useful if one goes beyond homology. It can be used to give a universal coefficient spectral sequence (due to Adams I think) for the (co)homology with coefficients in module spectrum over a ring spectrum. In general this spectral sequence does not degenerate to short exact sequences so the problem of splitting is not even (that) relevant. However, for for instance $K$-theory it does but I imagine (though I don't know but others certainly do) that even there one can find examples of non-splitting.</p>

http://mathoverflow.net/questions/27985/splitting-of-the-universal-coefficients-sequence/28538#28538 Robert Bruner 2010-06-17T17:33:22Z 2010-06-19T13:20:11Z

<p>(This is a corrected version of my original, off-the--cuff answer).</p> <p>If R has proj dim 1 and C is a flat chain complex, then you get the UCT sequence. To get the splitting, you need to also assume that C is projective, so that the map C -->> B from the complex to the subcomplex of boundaries has a splitting. </p> <p>A spectrum level construction of the splitting seems unlikely since you don't have kernels and cokernels, but only fibers and cofibers there.</p>