Classifying space for U(n): Difference between revisions

Revision as of 21:16, 15 February 2019

Revision as of 21:19, 15 February 2019

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There are homotopy fiber sequences

: <math> \mathbb{S}^{2n-1} \to B U(n-1) \to B U(n) </math>

Concretely, a point of the total space <math>BU(n-1)</math> is given by a point of the base space <math>BU(n)</math> classifying a complex vector space <math>V</math>, together with a unit vector <math>u</math> in <math>V</math>; together they classify <math> u^\perp < V </math> while the splitting <math>V = (\mathbb{C} u) \oplus u^\perp </math>, trivialized by <math>u</math>, realizes the map <math> B U(n-1) \to B U(n) </math> representing direct sum with <math>\mathbb{C}.</math>

Applying the [[Gysin Sequencesequence]], one has a long exact sequence

: <math> H^p ( BU(n) ) \overset{\smile d_{2n} \eta}{\tolongrightarrow} H^{p+2n} ( BU(n) ) \overset{j^*}{\tolongrightarrow} H^{p+2n} (BU(n-1)) \overset{\partial}{\tolongrightarrow} H^{p+1}(BU(n)) \dots </math>

where <math>\eta</math> is the [[fundamental class]] of the fiber <math>\mathbb{S}^{2n-1}</math>. By properties of the Gysin Sequence{{Citation needed|date=May 2016}}, <math>j^*</math> is a multiplicative homomorphism; by induction, <math>H^*BU(n-1)</math> is generated by elements with <math> p < -1 </math>, where <math>\partial</math> must be zero, and hence where <math>j^*</math> must be surjective. It follows that <math>j^*</math> must '''always''' be surjective: by the [[universal property]] of [[polynomial ring]]s, a choice of preimage for each generator induces a multiplicative splitting. Hence, by exactness, <math>\smile d_{2n}\eta </math> must always be '''injective'''. We therefore have [[short exact sequence]]s split by a ring homomorphism

: <math> 0 \to H^p ( BU(n) ) \overset{\smile d_{2n} \eta}{\tolongrightarrow} H^{p+2n} ( BU(n) ) \overset{j^*}{\tolongrightarrow} H^{p+2n} (BU(n-1)) \to 0 </math>

Thus we conclude <math>H^*(BU(n)) = H^*(BU(n-1))[c_{2n}]</math> where <math>c_{2n} = d_{2n} \eta</math>. This completes the induction.

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For the ''n''-torus, ''K''₀(B''Tⁿ'') is numerical polynomials in ''n'' variables. The map ''K''₀(B''Tⁿ'') → ''K''₀(BU(''n'')) is onto, via a [[splitting principle]], as ''Tⁿ'' is the [[maximal torus]] of U(''n''). The map is the symmetrization map

:<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