# Integral cohomology operations related to Landweber-Novikov

Let $U^* \rightarrow H^*$ be the homomorphism describing the complex orientation of $H^*$ from complex cobordism. Let $t_1, t_2, ...$ be indeterminates.

My question is: Does there exist an integral cohomology operation $H^* (X) \rightarrow H^*(X)[\mathbf{t}]$ that makes the following diagram commute for any $X$? If so can we describe it in a way that doesn't involve complex cobordism?

Here $s_{\mathbf{t}}$ is the total Landweber-Novikov operation.

In particular, since the Landweber-Novikov operation satisfies the Riemann-Roch type formula for proper, complex-oriented maps $$s_{\mathbf{t}} f_{*}x = f_{*}(c_{\mathbf{t}}(\nu_{f}) \cdot s_{\mathbf{t}}x)$$ (where $\nu_{f}$ is the virtual class $1-\nu_i$ and $\nu_i$ is the stable normal bundle of the proper, complex-oriented map $f: Z \rightarrow X$) we would expect ths operations to satisfy something similar.

EDIT: Whoops, original title didn't have to do with the question :) But it did have to do with the motivation behind asking it! Maybe another time...

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This follows from the answer to Integral cohomology (stable) operations, which asserts that the set of stable cohomology operations in integral cohomology of a given degree $2k>0$ is finite, and the observation that $s_{(k)}(cf_1) = cf_1^{k+1}$, where $cf_1\in U^2(BU)$ is the first universal Conner-Floyd-Chern class.
To see this, suppose that there is a stable operation $\sigma_{(k)}:H^\ast(-)\to H^{\ast+2k}(-)$ which extends the stable operation $s_{(k)}:U^\ast(-)\to U^{\ast+2k}(-)$. Then we would have $$\sigma_{(k)}(c_1)= c_1^{k+1},$$ where $c_1\in H^2(BU)$ is the first universal Chern class. But since the sum of stable operations is stable, we would have $$n\sigma_{(k)}(c_1)=0$$ for $n$ sufficiently large, contradicting the fact that $nc_1^{k+1}\neq 0$.
Alternatively, since $BU=K(\mathbb{Z},2)$, you can simply observe directly that the operation $x\mapsto x^{k+1}$ for $|x|=2$ does not extend to a stable operation. –  Eric Wofsey Dec 17 '12 at 20:46
Certainly $BU$ is not a $K(\mathbb{Z}, 2)$, Eric... It has all sorts of homotopy groups... –  Dylan Wilson Dec 17 '12 at 20:51
Oops, sorry, my brain is not functioning. I read that as $BU(1)$. –  Eric Wofsey Dec 17 '12 at 20:54
A variant of this argument also shows that you can't have an unstable operation: just look at $\Sigma c_1$ in $\Sigma BU$. Since all unstable operations on odd-degree classes are torsion, no unstable operation can take $\Sigma c_1$ to $\Sigma(c_1^{k+1})$. –  Eric Wofsey Dec 17 '12 at 20:56