Classifying space for U(n): Difference between revisions

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:<math>EU(n)=\left \{e_1,\ldots,e_n \ : \ (e_i,e_j)=\delta_{ij}, e_i\in \mathcal{H} \right \}.</math>
 
Here, ''H'' is an infinite-dimensional complex Hilbert space, the ''e''<sub>''i''</sub> are vectors in ''H'', and <math>\delta_{ij}</math> is the [[Kronecker delta]]. The symbol <math>(\cdot,\cdot)</math> is the [[inner product]] on ''H''. Thus, we have that EU(''n'') is the space of [[orthonormal]] ''n''-frames in ''H''.
 
The [[group action]] of U(''n'') on this space is the natural one. The [[base space]] is then
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where the <math>c_i</math>'s are the [[symmetric polynomials]] in the <math>x_i</math>'s.<math>\Box</math>
 
In contrast to the above description of <math>H^*(BU(n))</math>, many authors allow non-homogeneous elements in the cohomology, leading to the description <math>H^*(BU(n)) = \mathbb{Z}[[c_1,c_2,...,c_n]]</math>.<ref>Adams, 1974 p. 49</ref>.
 
==K-theory of BU(''n'')==
 
Let us consider topological complex K-theory as the cohomology theory represented by the spectrum <math>KU</math>. In this case, <math>KU^*(BU(n))\cong \mathbb{Z}[t,t^{-1}][[c_1,...,c_n]]</math>,<ref> Adams 1974, p. 49</ref>, and <math> KU_*(BU(n))</math> is the free <math>\mathbb{Z}[t,t^{-1}]</math> module on <math>\beta_0</math> and <math>\beta_{i_1}\ldots\beta_{i_r}</math> for <math>n\geq i_j > 0</math> and <math>r\leq n</math>.<ref> Adams 1974, p. 47</ref> In this description, the product structure on <math> KU_*(BU(n)) </math> comes from the H-space structure of <math>BU</math> given by Whitney sum of vector bundles. This product is called the [[Pontryagin product]].
 
{{warning| The following seems to be a computation of <math>KU_*BU(n)</math>, where <math>KU_*BU(n)</math> gets a ring structure from the tensor product H-space structure on <math>BU</math>. The statement needs clarification.}}
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:<math>f(w_1,\dots,w_n) \mapsto \frac{1}{n!} \sum_{\sigma \in S_n} f(x_{\sigma(1)},\dots,x_{\sigma(n)})</math>
 
and the image can be identified as the symmetric polynomials satisfying the integrality condition that
 
:<math> {n \choose n_1, n_2, \ldots, n_r}f(k_1,\dots,k_n) \in \mathbf{Z}</math>