Formal group law for oriented bordism

Question

From this answer I learned that the coefficient ring $MSO^{*}[1/2]$ of oriented bordism with 2 inverted supports an odd formal group law and is infact the universal such ring. Is there a reference/proof for this fact?

As motivation, I should mention that I'm trying to prove the fact that if $MU^{*}\rightarrow E^{*}$ is a map of rings where $E^{*}$ is a Landweber exact ring such that the image of the degree 2 generator $z_1\in MU^{*}\simeq \mathbb{Z}[z_1, \dots, ]$ vanishes in $E^{*}$, then this map must factor through the canonical "forget complex structure" map $MU^{*}\rightarrow MSO^{*}[1/2]$. If I can show that these conditions imply that $E^{*}$ has an odd formal group law, then I'd be able get my desired factorization $MSO^{*}[1/2]\rightarrow E^{*}$.

It therefore would be very helpful to understand why $MSO^{*}[1/2]$ itself has an odd formal group law (in particular the universal such one), as it might be the case that the proof method for $MSO^{*}[1/2]$ uses only that $MSO^{*}[1/2]$ is a Landweber exact ring for which the image of $z_1$ vanishes under the map $MU^{*}\rightarrow MSO[1/2]^{*}$ and hence generalizes straightaway to my general case.

Note: An odd (one dimensional) formal group law $F(X, Y)$ is one for which $F(X, -X)=0$.

Answer 1

One basic point is as follows. The usual inclusions $U(n)\to O(2n)$ give rise to a map $BU\to BSO$ of $E_\infty$ spaces and then a map $MU\to MSO$ of ring spectra, which gives a complex orientation of $MSO$. Note that a map $X\to MSO(n)$ of spaces represents a class in $MSO^n(X)$. The orientation class $x\in MSO^2(\mathbb{C}P^\infty_+)=MSO^2(BSO(2)_+)$ is just the zero section $BSO(2)_+\to BSO(2)^T=MSO(2)$, where $T$ is the tautological bundle. There is a suspension isomorphism $MSO^2(BSO(2)_+)=MSO^3(\Sigma BSO(2)_+)$ and $\Sigma x$ is represented by an evident map $\Sigma BSO(2)_+\to MSO(3)$, covering $Bi$, where $i$ is the evident inclusion $i\colon SO(2)\to SO(3)$. It is a key point that $i$ extends to give a map $O(2)\to SO(3)$, with $i(A)=A\oplus\det(A)$. Covering this, we get a map $BO(2)^{\lambda ^2T}\to MSO(3)$ extending $\Sigma x$. Now choose $g\in O(2)\setminus SO(2)$, and let $\beta$, $\gamma$ and $\delta$ be the automorphisms of $SO(2)$, $O(2)$ and $SO(3)$ given by conjugating with $g$, $g$ and $i(g)$ respectively. As $\gamma$ and $\delta$ are inner automorphisms, it is standard that $B\gamma$ and $B\delta$ are homotopic to the respective identity maps, and one can check that they also induce maps of the relevant Thom spaces that are homotopic to the identity. From this it follows that $\beta^*(\Sigma x)=\Sigma x$. However, $\beta$ reflects the suspension coordinate and acts as the inversion map on $SO(2)$ so we have $\beta^*(\Sigma x)=-\Sigma [-1](x)$. As this is equal to $\Sigma x$, we see that the formal group law is odd.

I am sure that I have seen the oddness proved somewhere in the literature, but I do not remember where. I thought it was in a paper by Steve Mitchell, but if so, I have not found the right one. I think that the argument given was not the same as the one above, but I do not remember what it was.