We can regard an oriented manifold as an unoriented manifold by forgetting the orientation. This gives a homomorphism from the oriented cobordism group to the unoriented cobordism group. What is the image of this homomorphism? For example, in 4 dimensions, the unoriented cobordism group is $Z_2 \times Z_2$, the oriented cobordism group is $Z$ and is generated by $CP^2$, and the image of the homomorphism is $Z_2$. Is it known what this image is in arbitrary dimensions?
1 Answer
I will write $MO_*$ for the unoriented cobordism ring and $MSO_*$ for the oriented cobordism ring. Inside of $MO_*$, there is a subring $W$ defined by C. T. C. Wall, which is the cobordism ring of $w_1$-spherical manifolds, i.e., compact smooth manifolds $M$ whose first Stiefel-Whitney class $w_1\in H^1(M; F_2)$ lifts to a class in integral cohomology $H^1(M; Z)$. (Recall that $w_1$ is zero for an oriented manifold, so being $w_1$-spherical is weaker than being oriented.)
Algebraically: recall that Thom proved in "Quelques propriétés globales des variétés différentiables" (1954) that $MO_*$ is isomorphic to a polynomial algebra $F_2[x_2, x_4, x_5, x_6, x_8, x_9, ...]$ on a single generator $x_n$ in each degree $n$ which is not one less than a power of $2$. Wall, in "Determination of the cobordism ring" (1960), proved that if you choose the generators $x_2, x_4, ...$ of $MO_*$ carefully, then $W$ is the subring of $MO_*$ generated by
- $x_{2n}$, for each $n$ not a power of $2$,
- $x_{2n-1}$, for each $n$ not a power of $2$,
- and $x_{n}^2$, for each $n$ which is a power of $2$.
In "Determination of the cobordism ring," Wall considers a derivation $\delta: W \rightarrow W$ given by $\delta(x_{2n}) = x_{2n-1}$, and $\delta(x_{2n-1}) = 0$, and $\delta(x_n^2) = 0$. Wall proves that the image of the map $MSO_* \rightarrow MO_*$ is precisely $\ker \delta \subseteq W \subseteq MO_*$.
You can do a bit of calculation by hand, of the kernel of the derivation $\delta$ on a polynomial $F_2$-algebra with the generators I described, to see what the image of $MSO_* \rightarrow MO_*$ looks like explicitly in various degrees. You know from the description I just gave that whatever you get, it will be contained in the subalgebra $W\subseteq MO_*$. Here is another nice fact: whatever you get, it will contains all the elements of $MO_*$ which are squares, by the theorem from Milnor's 1963 "On the Stiefel-Whitney numbers of complex manifolds and spin manifolds" which states that the image of the map $MU_* \rightarrow MO_*$ is precisely all the squares in $MO_*$. Here $MU_*$ is the complex cobordism ring, and of course $MU_* \rightarrow MO_*$ factors through $MSO_* \rightarrow MO_*$.
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1$\begingroup$ I think a little fiddling with Wal's result stated as above shows that this image is precisely the subalgebra of squares. $\endgroup$ Commented Oct 27 at 17:41