While reviewing the computation of the complex bordism ring $\pi_*MU$, I managed to masterfully tangle myself up in some very confusing indexing issues in co/homology. For context, this happened in the step of computing the cohomology of $MU$ (before obtaining its homotopy through the Adams spectral sequence). In this post I describe the cohomology part of the computation, and the subtlety I encountered while going through it.

Cohomology of $BU(n)$

The input to our computation is the cohomology of $BU(n)$, which is usually computed by induction on $n$, though I’ll just tell you what the final result is. To state it we consider the homomorphism $U(1)^n \to U(n)$ which just embeds the $n$ entries as a diagonal matrix. This gives maps on the classifying spaces $BU(1)^n \to BU(n)$.

Theorem: The induced map on cohomology $E^*BU(n) \to E^*BU(1)^n$ is injective, and its image is precisely the subalgebra of elements invariant under permutations of the $n$ factors.

Since $BU(1) \simeq \mathbb{CP}^\infty$, the integral cohomology $E^*\mathbb{CP}^\infty \cong (\pi_*E)[[t]]$ is known – $E$ here is a complex oriented cohomology theory and $t$ is a choice of Thom orientation element for it sitting in degree $2$ of the grading. Hence $E^*BU(1)^n \cong (\pi_*E)[[t]]^{\otimes n} = (\pi_*E)[[t_1,\cdots,t_n]]$. Consequently, we can identify $E^*BU(n)$ with the algebra of symmetric polynomials (or rather, formal power series), which is also itself a polynomial algebra generated by the elementary symmetric polynomials. The $i$th elementary symmetric polynomial on $t_1,\cdots,t_n$ is typically denoted $c_i$, and it represents the $i$th Chern class.

\[E^*BU(n) \cong (\pi_*E)[[c_1,\cdots,c_n]]\]

Before translating this to the associated Thom spectra $MU(n)$, let’s take a minute to remind ourselves about the Thom construction in general.

Thom constructions

There’s a few decent notations out there for the Thom construction, but I tried coming up with my own notation that emphasizes its role as a “twisted suspension”:

Definition: If $Y$ is a space and $\xi$ is a sphere bundle over it, then the Thom space of $\xi$ is the homotopy cofiber of the bundle projection:

\[\Sigma_+^\xi Y := \operatorname{hocofib}(\xi \to Y)\]

Most commonly, this is applied on sphere bundles appearing as unit spheres in vector bundles: so in a real vector bundle of rank $n$ you’re gonna get a bundle of $S^{n-1}$ spheres, and in a complex vector bundle of rank $n$ you’d get $S^{2n-1}$.

If, for example, we denote by $\underline{n}$ the trivial bundle $\mathbb{R}^n\times Y \to Y$, or the associated sphere bundle $S^{n-1}\times Y \to Y$, then we recover the usual $n$-fold suspension

\[\Sigma_+^{\underline{n}}Y \simeq \Sigma_+^n Y\]

This can be proved easily in the special case $Y \simeq pt$ using pasting of the following homotopy pushouts:

For the general case, note that smashing with $Y$ preserves homotopy colimits.

Arguably the most important feature of co/homology is that interacts well with colimits, and in particular “preserves suspensions” with the cost of shifting its indexing

\[E^*(\Sigma^n Y) \cong E^{*-n}(Y)\]

Even though we subtracted $n$ from the degree, the grading of the right-hand side is actually raised by $n$; degree $n$ of $E^{*-n}$ is degree zero of $E^*$ and so on. To reflect this, for any graded object $A = A^*$ I shall denote the shifted object by $A[n]^*$

\[A[n]^* := A^{*-n}\]

Another minor adjustment we should consider is using reduced cohomology $\widetilde{E}^*$ which recovers $E^*$ upon adjoining a basepoint (that’s what the little $+$ subscript indicates). In this form, the suspension isomorphism says

\[\widetilde{E}^*\Sigma^n_+Y \cong (E^*Y)[n]\]

But I’ll keep this notation for later. In contrast, the Thom isomorphism says, for any $S^{n-1}$-bundle $\xi$ over $Y$:

\[\widetilde{E}^*\Sigma_+^\xi Y \cong E^{*-n}Y\]

Now you can see why I chose my notation. There’s a few good ways to prove this identity; I think the most conceptual one is just by proving it locally. Indeed, any bundle is locally trivial so the Thom space is locally a suspension. The Thom isomorphism is usually stated more specifically, by saying that the isomorphism, from right to left, is defined as multiplication with the Thom class $t$ of $\xi$. Hence it may be more accurate to write

\[\widetilde{E}^*\Sigma_+^\xi Y \cong tE^{*-n}Y\]

Since $t$ has degree $n$, multiplying with it increases the total degree of the RHS from $*-n$ to $*$, making it compatible with the LHS. In this form then, it is not just an isomorphism of modules, but of graded modules. Another equivalent way of writing this is obtained by just shifting some indices further:

\[\widetilde{E}^{*+n}\Sigma_+^\xi Y \cong t E^*Y\]

This form is what we will use shortly.

The spectra $MU(n)$

For each $n$, the Thom spectrum $MU(n)$ is defined as follows

\[MU(n) := \Sigma_+^{\infty + \gamma_n - 2n}BU(n)\]

where $\gamma_n$ is the tautological bundle over $BU(n)$. One also denotes $MU := \operatorname{colim}_{n\to\infty}MU(n)$ though I won’t use this in the rest of the post.

Observe that I’m allowing myself to mix three different kinds of operations here in one symbol:

  • $\Sigma^\infty$ sends a pointed space to its suspension spectrum.
  • $\Sigma^{\gamma_n}_+$ is our Thom construction.
  • $\Sigma^{-2n}$ is the usual suspension functor – actually, with the minus sign, that is a desuspension which has to be performed only after applying $\Sigma^\infty$.

The desuspension by $2n$ is indented to counterbalance $\gamma_n$ which is $n$-dimensional over $\mathbb{C}$ and $2n$-dimensional over $\mathbb{R}$. In fact you can equivalently consider the virtual vector bundle $\gamma_n - \underline{2n}$ which has virtual dimension $0$. This equivalence perhaps justifies my abuse of notation.

It’s possible to express the Thom spectra $MU(n)$ a bit more concretely: it shouldn’t be too tough to convince yourself that restricting the tautological bundle $\gamma_n$ along $BU(n-1) \hookrightarrow BU(n)$ yields $\gamma_{n-1}\oplus\underline{1}$. This has a striking consequence: the sphere bundle associated to $\gamma_n \to BU(n)$ is equivalent to the subgroup inclusion $BU(n-1) \hookrightarrow BU(n)$! This inclusion is a cofibration so the homotopy cofiber in the definition of the Thom space simply becomes a mere quotient. In symbols:

\[\Sigma_+^{\gamma_n}BU(n) \simeq BU(n)/BU(n-1)\]

or, phrased in terms of Thom spectra,

\[MU(n) \simeq \Sigma^{\infty-2n}BU(n)/BU(n-1)\]

At this point we have two ways to approach the cohomology of $MU(n)$: either use the Thom isomorphism directly, or use the explicit form that we just found. Of course both ways are equivalent and they should yield the same result, but if you’re not careful the results can seem slightly different.

1 - Via the Thom isomorphism

We defined $MU(n)$ as the suspension spectrum of $\Sigma_+^{\gamma_n}BU(n)$, but desuspended $2n$ times. Here $\gamma_n$ is the tautological bundle of rank $2n$, whose Thom class is the $n$th chern class $c_n$. The cohomology of a suspension spectrum is the reduced cohomology of the space, and applying the suspension isomorphism in reverse reveals that desuspending amounts to increasing the index:

\[E^*MU(n) = \widetilde{E}^{*+2n}\Sigma_+^{\gamma_n}BU(n)\]

This puts us in a perfect position to apply the Thom isomorphism, which yields

\[\boxed{E^*MU(n) \cong c_n E^*BU(n)}\]

Easy, right? But now let’s look at the other approach.

2 - Via a long-exact sequence

We have established an equivalence of spectra $MU(n) \simeq \Sigma^{\infty-2n}BU(n)/BU(n-1)$. As with any cofiber, the cohomology of the RHS fits into a long exact sequence,

\[\cdots \longrightarrow E^{*+2n}BU(n)/BU(n-1) \longrightarrow E^{*+2n}BU(n) \longrightarrow E^{*+2n}BU(n-1) \longrightarrow \cdots\]

In fact you could prove that this segment of the long exact sequence breaks off into a short exact sequence. Therefore $E^{*+2n}BU(n)/BU(n-1)$ is an ideal in $E^{*+2n}BU(n)$, namely the kernel of the map on the right. That map is induced by the inclusion $BU(n-1) \hookrightarrow BU(n)$ – in terms of chern classes this can be identified as

\[(\pi_*E)[[c_1,\cdots,c_n]] \xrightarrow{\qquad\qquad} (\pi_*E)[[c_1,\cdots,c_{n-1}]]\]

sending $c_n$ to $0$ and the other generators to themselves. We see that its kernel is precisely the ideal generated by $c_n$:

\[\boxed{E^*MU(n) \cong c_nE^{*+2n}BU(n)}\]

Reconciliation

We got two different answers, so what went wrong? The issue is a small subtlety in how the Thom isomorphism shifts degrees – what we should really do is first multiply by the Thom class, and then shift everything down. In the improved notation for shifts, what I’m saying is

\[c_n (E^*BU(n)[-2n]) \not\cong (c_n E^*BU(n))[-2n]\]

The latter of these is the correct interpretation.

So that’s it. This is the punchline to this whole story. Might seem a bit anticlimactic, but, well, that didn’t stop me from getting stuck on it for a few good hours.