# Chromatic Homotopy Theory – I

In this post, I will talk about chromatic homotopy theory, which you can call “chromotopy”. This post assumes familiarity with spectra and the notions of Chern classes of line bundles; the reader who is not familiar with this notion is referred to (the very brief) http://www.mit.edu/~sanathd/papers/htpy.pdf.

Recall that there is a universal line bundle $\mathcal{O}(1)$ over $BU(1)$ such that every line bundle over a space $X$ can be obtained by pulling back over maps $X\to BU(1)$. The first Chern class is of interest in this situation. This satisfies $c_1(f^\ast \mathscr{L}) = f^\ast c_1(\mathscr{L})$ where $\mathscr{L}\to X$ is a line bundle over $X$ and $c_1(\mathscr{L})\in H^2(X;\mathbf{Z}) = H^2(X)$. Since $BU(1)$ acts as a moduli space for complex line bundles, this means that the Chern class of any line bundle is determined by $c_1(\mathcal{O}(1))$, which lies in $H^2(\mathbf{CP}^\infty)$. Recall that a property of the Chern class is that $c_1(\mathscr{L}\otimes \mathscr{L}^\prime) = c_1(\mathscr{L}) + c_1(\mathscr{L}^\prime)$. How can we specify the Chern class purely in terms of cohomology?

We claim that $H^\ast(\mathbf{CP}^\infty)\cong\mathbf{Z}[[t]]$ where $t\in H^2(\mathbf{CP}^\infty)$, i.e., $t$ is of degree $2$. To see this, use the Serre spectral sequence (we refer the reader to http://www.mit.edu/~sanathd/papers/spectral%20sequences.pdf for very brief notes). Consider the fibration $S^1\to S^{2n+1}\to \mathbf{CP}^n$. The Serre spectral sequence then reads $H^s(\mathbf{CP}^n;H^t(S^1))\Rightarrow H^{s+t}(S^{2n+1})$, where we are implicitly assuming coefficients in $\mathbf{Z}$. Now, the group $H^s(\mathbf{CP}^n;H^t(S^1))$ is $H^s(\mathbf{CP}^n)$ if $t=0,1$, and is zero otherwise. Hence, the $E_2$ page looks like:

Where the only nontrivial differentials are shown above. Since $H^k(S^{2n+1})$ is $\mathbf{Z}$ if $k=0,2n+1$, and is zero otherwise, and $H^1(\mathbf{CP}^n)$ lies on the line $s+t = 1$, we see that $H^1(\mathbf{CP}^n) \cong 0$. From this it follows that $H^{2k+1}\mathbf{CP}^n \cong 0$ for all $k$. Clearly via $d_2$ each even cohomology group of $\mathbf{CP}^n$ must be isomorphic, so it suffices to determine $H^0\mathbf{CP}^n$. But this is simply $\mathbf{Z}$, finishing our computation of the cohomology groups of $\mathbf{CP}^n$. The ring structure follows from the multiplicative structure on the spectral sequence, so $H^\ast(\mathbf{CP}^n)\cong \mathbf{Z}[x]/(x^{n+1})$; taking the limit finishes the proof.

Using this computation, one can compute the $E$-cohomology of $\mathbf{CP}^\infty$ for more general cohomology theories $E$. Let $E$ be a multiplicative cohomology theory (so that $E^\ast(X)$ has the structure of a graded ring for every $X$; this is equivalent to asking that (the spectrum associated via Brown representability to) $E$ is a homotopy ring spectrum). Say that $E$ is a complex-orientable cohomology theory if the Atiyah-Hirezebruch spectral sequence $E^{p,q}_2 = H^p(X;E^q(\ast))\Rightarrow E^{p+q}(X)$ degenerates on the second page. Then clearly $E^\ast(\mathbf{CP}^\infty) \cong E^\ast(\ast)[[t]]$ where $t\in E^2(\ast)$. A complex-orientable cohomology theory with a choice of $t\in E^2(\ast)$ is called complex-oriented. With a similar computation, we see that $E^\ast(\mathbf{CP}^\infty\times\mathbf{CP}^\infty)\cong E^\ast(\ast)[[x,y]]$, where $x$ and $y$ are the pullbacks of $t$ along the projections $\pi_1,\pi_2:\mathbf{CP}^\infty\times\mathbf{CP}^\infty\to\mathbf{CP}^\infty$.

In ordinary cohomology, we can define $t$ to be the first Chern class $c_1(\mathcal{O}(1))$. Then the Chern class of any line bundle $\mathscr{L}\to X$ is the pullback in cohomology of $t$. Is this consistent with the Chern class preserving tensor products? By Yoneda’s lemma, the tensor product of line bundles can be thought of as a multiplication $\mathbf{CP}^\infty\times\mathbf{CP}^\infty\to\mathbf{CP}^\infty$ inducing $\mu:H^\ast(\mathbf{CP}^\infty)\to H^\ast(\mathbf{CP}^\infty\times\mathbf{CP}^\infty)$. But this is simply a map $\mu:\mathbf{Z}[[t]]\to\mathbf{Z}[[x,y]]$. Consider $\mu(t)$; this is just $\mu(c_1(\mathcal{O}(1))) = c_1(\mu(\mathcal{O}(1))) = c_1(\pi_1^\ast(\mathcal{O}(1))\otimes\pi_2^\ast(\mathcal{O}(2)))$. On the other hand, $x+y = \pi_1^\ast(t) + \pi_2^\ast(t) = c_1(\pi_1^\ast(\mathcal{O}(2))) + c_1(\pi_2^\ast(\mathcal{O}(2)))$. The formula $\mu(t) = x+y$ gives the formula $c_1(\mathscr{L}\otimes \mathscr{L}^\prime) = c_1(\mathscr{L}) + c_1(\mathscr{L}^\prime)$.

What about the more general case of a complex-oriented cohomology theory? In this case, denote by $c_1^E(\mathscr{L})$ the cohomology class $f^\ast t\in E^2(X)$ where $f:X\to\mathbf{CP}^\infty$. We see that

$c_1^E(\pi_1^\ast(\mathcal{O}(1))\otimes\pi_2^\ast(\mathcal{O}(2))) = f(c_1(\pi_1^\ast(\mathcal{O}(2))), c_1(\pi_2^\ast(\mathcal{O}(2)))) = f(x,y)\in E^\ast(\mathbf{CP}^\infty\times\mathbf{CP}^\infty)$

for some function $f(x,y)$. Consequently, $c_1^E(\mathscr{L}\otimes\mathscr{L}^\prime) = f(c_1^E(\mathscr{L}),c_1^E(\mathscr{L}^\prime))$. This function $f(x,y)$ clearly satisfies conditions reflecting the geometry of line bundles, namely:

• Let $\mathscr{O}$ denote the trivial line bundle; then $\mathscr{L}\otimes\mathscr{O}\simeq\mathscr{L}\simeq\mathscr{O}\otimes\mathscr{L}$, so $f(x,0) = x = f(0,x)$.
• $\mathscr{L}\otimes\mathscr{L}^\prime\simeq\mathscr{L}^\prime\otimes\mathscr{L}$, so $f(x,y) = f(y,x)$.
• $(\mathscr{L}\otimes\mathscr{L}^\prime)\otimes\mathscr{L}^{\prime\prime}\simeq\mathscr{L}\otimes(\mathscr{L}^\prime\otimes\mathscr{L}^{\prime\prime}$, so $f(f(x,y),z) = f(x,f(y,z))$.

If you have studied elliptic curves (or number theory, or algebraic geometry, etc.), you’ll notice these relations are exactly those of a formal group law. What this shows is the following:

Proposition: A complex-oriented cohomology theory $E$ gives a formal group law over $E^{2\ast}(\ast)$.

What are some examples of this? As we saw above with ordinary cohomology, the formal group law that arises is the additive formal group law $f(x,y) = x+y$. Suppose $E = KU$, complex K-theory. This is complex-oriented. Without proof, we claim that the Chern class of a line bundle $\mathscr{L}\to X$ is $1-\mathscr{L}\in KU^\ast(X)$. Then $c_1(\mathscr{L}\otimes\mathscr{L}^\prime) = 1-\mathscr{L}\otimes\mathscr{L}^\prime$. This is:

$2-\mathscr{L}-\mathscr{L}^\prime-1+\mathscr{L}+\mathscr{L}^\prime-\mathscr{L}\otimes\mathscr{L}^\prime = (1-\mathscr{L}) + (1-\mathscr{L}^\prime) - (1-\mathscr{L})(1-\mathscr{L}^\prime) = c_1(\mathscr{L}) + c_1(\mathscr{L}^\prime) - c_1(\mathscr{L})c_1(\mathscr{L}^\prime).$

In other words, the formal group law for $KU$ is $f(x,y) = x+y-xy$, which can easily be checked to satisfy the conditions for being a formal group law. Consider a formal group law $f(x,y) = \sum_{i,j}c_{ij}x^iy^j$. Then a formal group law over a ring $R$ can be considered to be a map $\mathbf{Z}[c_{ij}]/I\to R$ where $I$ is the ideal generated by the relations for being a formal group law (in terms of $c_{ij}$), namely:

• $c_{i0} = 1$ if $i=1,0$, and $0$ otherwise, and similarly for $c_{0i}$.
• $c_{ij} = c_{ji}$ for all $i,j$.
• Some complicated relations, not reproduced here, corresponding to associativity.

Hence there is a univeral formal group law over $L = \mathbf{Z}[c_{ij}]/I$ such that every formal group law over a ring $R$ can be obtained by pushing the universal formal group law to $R$. Consequently, a complex-oriented cohomology theory $E$ has a map $L\to E^\ast$. Lazard proved the following important result.

Proposition: The Lazard ring $L$ is a polynomial algebra $\mathbf{Z}[x_1,x_2,\cdots]$ such that there is a grading on $L$ with $x_i$ of degree $-2i$.

Shifting over to topology, we see that there is a universal example of a complex-oriented spectrum. Define $MU$ to be the Thom spectrum of $BU$, defined as the homotopy colimit of the sequence $MU(0)\to MU(1)\to MU(2)\to\cdots$ where $MU(n)$ is the spectrum $\Sigma^{\infty-2n}BU(n)/BU(n-1)$. For example, $MU(0)$ is the sphere spectrum and $MU(1) = \Sigma^{\infty-2}\mathbf{CP}^\infty$. The universal bundle of rank $n$ on $BU(n)$ is classified by a map $MU(n)\to E$. In particular, the map $MU(1)\to E$ is given by the complex orientation of $E$ (because $[MU(1),E] = \widetilde{E}^2(BU(1))$). These maps are compatible in the sense that the following diagram commutes for $k:

Consequently we get a map $MU\to E$, which is a map of ring spectra. In other words, we have:

Proposition: Consider the inclusion $MU(1)\to MU$, giving a complex orientation $t$ of $MU$. Then the map taking $[MU,E]$ to the set of complex orientations of $E$ given by taking $\phi:MU\to E$ to $\phi(t)$ is a bijection.

Therefore $MU$ is the universal complex oriented spectrum. Milnor and Novikov proved:

$MU_\ast\cong\mathbf{Z}[x_1,x_2,\cdots]$

with $\dim x_i = 2i$. Now, $MU^\ast$ is isomorphic to $MU_\ast$ with the grading reversed – but this looks suspiciously like the Lazard ring. Quillen bridged this topological side with algebra in the following result.

Theorem: The map $L\to MU^\ast$ classifying the formal group law of $MU$ is an isomorphism.

Now, given an $L$-module $R$, we can attempt to define a new cohomology theory via $E^\ast(X) = MU^\ast(X)\otimes_L R$, but this does not necessarily have excision or Mayer-Vietoris. Therefore there needs to be some conditions on the map $L\to R$ making $E$ a cohomology theory. To do this, we must use the language of stacks. We’ll pick up from here in the next post (but if you’re interested, the next part of the story is written down here: http://www.mit.edu/~sanathd/papers/chromatic%20homotopy%20theory.pdf).