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I have some questionss about the construction of the complex bordism ring MU and would appreciate every answer:

  1. I have read that the multiplication in MU is given by the tensor product of vector bundles $BU(n) \times BU(m) \rightarrow BU(n+m)$. How does this tensor product yield us a map $MU(n) \wedge MU(m) \rightarrow MU(n+m) $, I do not get how we can go from $BU$ to $MU$?

  2. What is the unit for the multiplication on MU?

I would be happy, if you could give me an answer.

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To complement/summarize Tyler's answer: $MU$ is the Thom spectrum of the universal (0-dimensional virtual) complex vector bundle over $BU$, and its product map is induced from the map $BU \times BU \rightarrow BU$ guaranteed by the external tensor product with itself of the universal bundle. (But this only works up to homotopy. Presumably you could use the $E_\infty$-structure on $BU$ to tighten things up...) –  Aaron Mazel-Gee Jul 18 '12 at 9:26

1 Answer 1

The multiplication is actually induced by direct sum, or equivalently by block sum of matrices $U(n) \times U(m) \to U(n+m)$. (Tensor product lands in $U(nm)$.)

One definition of $MU(n)$ is as follows. Over $BU(n)$ we have a universal bundle $EU(n) \times \mathbb{C}^n / U(n)$, and the Thom space is obtained by taking the one-point compactification in each fiber and then identifying all the new points.

From this point of view, there is a natural map $$ EU(n) \times \mathbb{C}^n \times EU(m) \times \mathbb{C}^m \to EU(n+m) \times \mathbb{C}^{n+m} $$ that uses block sum on $EU$ and collects together the vector space factors. The map you're looking for comes from taking this map, passing to the quotient by the action of the unitary groups, and then adding this extra point on both sides (which turns the cartesian product into a smash product).

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