It is known that the $n^{\text{th}}$ Morava $K$-theory at a prime $p$, denoted $K(n)$, is complex oriented. In other words, it admits a theory of Chern classes, or equivalently a morphism of homotopy associative, commutative ring spectra $f:MU\to K(n)$. This complex orientation is related to the height $n$ Honda formal group law in the following way: the complex orientation of $K(n)$ tells us that $K(n)^\ast(\mathbb{C}P^\infty)\simeq \pi\_\ast(K(n)
)[[x]]$. Recall that there is an element in $x\_{MU}\in MU^2(\mathbb{C}P^\infty)$ such that, for the multiplication map $t:\mathbb{C}P^\infty\times\mathbb{C}P^\infty\to\mathbb{C}P^\infty$, $t^\ast(x\_{MU})$ is given by the universal formal group law $F\_u\in MU\_\ast[[x,y]]$. Then our map $f:MU\to K(n)$ maps $x\_{MU}$ to some $x\_{K(n)}\in K(n)^2(\mathbb{C}P^\infty)$. Then we obtain a formal group law by looking at $t^\ast(x\_{K(n)})\in K(n)^\ast(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)$ and choosing an isomorphism $$K(n)^\ast(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)\cong \pi\_\ast(K(n))[[x,y]].$$ If we choose *any* multiplicative map $MU\to K(n)$, will the above construction return the height $n$ Honda formal group law? If not, is it somehow important how we construct $K(n)$ to make sure that the right orientation shows up? Ultimately I'm interested in the moduli space of complex orientations on $K(n)$ and how this relates to formal group laws on $\pi\_\ast(K(n))$. Also, would changing the complex orientation (and thus potentially the construction of $K(n)$) affect which $A\_\infty$-structure on $K(n)$ with which we end up?



Uniqueness of Complex Orientation of Morava K-theory
First, I claim that if we ignore the ring structure and complex orientation, then there is a unique spectrum (up to homotopy equivalence) that deserves to be called $K(n)$. I do not know whether there is a nice way to see this before setting up a lot of chromatic homotopy theory based on a particular choice of $K(n)$. However, once one has set all that up, one can say, for example, that $K(n)$ is the unique spectrum $X$ such that $X=L\_nX$ and $X$ is a retract of $X\wedge F$ for some finite spectrum $X$ of type $n$ and $\pi\_\*(X)\simeq K(n)\_\*$ as graded abelian groups.


Now, there are various constructions that produce different versions of $K(n)$ as ring spectra or $MU$-algebra spectra, with or without an $A\_\infty$ structure. Because of the previous paragraph, these can be thought of as giving different ring structures, algebra structures or $A\_\infty$ structures on the same object.


Now suppose we have a formal group law $F$ of height $n$ over the ring $K(n)\_\*$, which is compatible with the grading in the usual way. I claim that the standard $K(n)$ admits a ring structure and orientation for which the associated FGL is $F$. Indeed, $F$ gives a ring map $\phi\_F\colon MU\_\*\to K(n)\_\*$. By thinking about the $p$-series and the grading we see that this must be surjective in nonnegative degrees. Now define $I\_F$ to be the kernel of $\phi\_F$, and define $v\_n$ to be the coefficient of $x^{p^n}$ in the standard FGL over $MU\_\*$. We then find that $\phi\_F$ induces an isomorphism $v\_n^{-1}MU\_\*/I\_F\to K(n)\_\*$. One can check that $I\_F$ is automatically generated by a regular sequence. At least at odd primes, it follows that there is a commutative ring object $K\_F$ in the category of strict $MU$-modules with $\pi\_\*(K\_F)\simeq v\_n^{-1}MU\_\*/I\_F$, and this is unique up to unique isomorphism in the relevant category. The underlying spectrum of $K\_F$ is equivalent to the standard $K(n)$, and the claim follows. Slightly weaker statements are still available if $p=2$.


One can also ask about the situation where we fix a ring structure on $K(n)$, and look at all the FGLs that we get from maps $MU\to K(n)$ that are compatible with that ring structure. Using the Atiyah-Hirzebruch spectral sequence we see that there exist elements $x\in K(n)^2\mathbb{C}P^\infty$ that restrict to the standard generator in $K(n)^2\mathbb{C}P^1$, and it follows in the usual way that there is at least one ring map $MU\to K(n)$, giving a formal group law $F$ say. After fixing this ring map, a standard line of argument shows that the full set of ring maps $MU\to K(n)$ bijects with the set of $K(n)\_\*$-algebra maps $K(n)\_\*MU\to K(n)\*$, and thus with the set of formal power series over $K(n)\_\*$ of the form $x+O(x^2)$. It follows in turn that the full set of possible FGLs that we can get is just the set of FGLs that are strictly isomorphic to $F$. In particular, if we set things up so that $F$ is the Honda FGL, then by changing the orientation we can get all of the uncountably many FGLs that are strictly isomorphic to the Honda FGL, but not equal to it.


I do not think that there is any known construction of a specific $A\_\infty$ structure on $K(n)$. The original theorem of Alan Robinson just showed that there were uncountably many structures, without specifying a particular one. I think that that theorem did not say anything about the map $MU\to K(n)$ being $A\_\infty$. I am not sure what there is in more recent papers.


Anyway, I think that the main message is that it does not matter very much which FGL you use, so long as it has height $n$. There are some minor technical advantages in using a $p$-local FGL so that you can exploit the fact that $BP\_\*$ is much sparser than $MU\_\*$. Whichever FGL you use on $K(n)\_\*$, you can always use the universal deformation to get a compatible version of $\widehat{E(n)}$. A theorem of Ando says that this $\widehat{E(n)}$ has a preferred coordinate that lifts the chosen coordinate on $K(n)$ and interacts nicely with power operations; but this works perfectly well for any choice of $K(n)$ coordinate.


[UPDATE: I previously attributed the $A\_\infty$ structure on $K(n)$ to Andy Baker, but I misremembered that; I have corrected it to Alan Robinson. See also Lennart Meier's comment: Vigleik Angeltveit has shown that the $A\_\infty$ structure is unique in a sense different from that studied by Robinson.]


