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The Atiyah-Hirzebruch spectral sequence is a strong computational tool that yields several interesting computation in (co)homology. I want to know whether $K_\ast(MU)$ and $K^\ast(MU)$ have been identified. Please redirect me to the relevant papers/books/articles, etc.

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2 Answers 2

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You can learn about this, and more, in Part II of Adams' "Stable Homotopy and Generalised Homology" (1974), especially section 4, in the special case $E = KU$.

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For a more self-contained answer, let $L$ be the tautological line bundle over $\mathbb{C}P^\infty$. This gives a class $x=[L]-[1]\in\widetilde{K}^0(\mathbb{C}P^\infty)$. It is standard that $K^0(\mathbb{C}P^\infty)$ is the formal power series ring $\mathbb{Z}[\![x]\!]$, and $K^0(\mathbb{C}P^n)=\mathbb{Z}[x]/x^{n+1}$. It follows that $K_0(\mathbb{C}P^n)$ is a free abelian group on classes $\beta_0,\dotsc,\beta_n$ with $\langle\beta_i,x^j\rangle=\delta_{ij}$, and then that $\widetilde{K}_0(\mathbb{C}P^\infty)$ is freely generated by $\\{\beta_i:i>0\\}$. Next, from the definition of the spectrum $MU$ there is a canonical map $f\colon\Sigma^{-2}\mathbb{C}P^\infty\to MU$. By Bott periodicity we can identify $K_0(\Sigma^{-2}\mathbb{C}P^\infty)$ with $\widetilde{K}_0(\mathbb{C}P^\infty)$ and then define $b_i=f_*(\beta_{i+1})\in K_0(MU)$. It can then be shown that $b_0=1$ and that $K_0(MU)$ is the polynomial ring $\mathbb{Z}[b_1,b_2,b_3,\dotsc]$. We also have $K_1(MU)=0$ and everything is $2$-periodic by Bott periodicity so $K_{*}(MU)=\mathbb{Z}[u,u^{-1},b_1,b_2,b_3,\dotsc]$ with $|u|=2$ and $|b_i|=0$. Also, $K^0(MU)$ is just the dual group $\text{Hom}(K_0(MU),\mathbb{Z})$.

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