In my recent research, I need to know if the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ are always nontrivial in the unoriented and oriented bordism rings for $n>2$. (For the motivation, by Benny Cheng, the cones over them are special Lagrangians in $\mathbb{C}^{n^2+n}$. I want to know if the cones are topologically smoothable. The $n=3$ case comes from Haskins & Pacini.)

I know that for $n=3$ we are dealing with the Wu manifold, and there are many references here. For $n>3$, I tried to find references about the Stiefel-Whitney numbers of these, which by Thom, is enough for the unoriented case. However, I've only been able to locate the cohomology ring of these in Topology of Lie Groups I & II by Mimura and Toda. I don't know much algebraic topology, but I think they did not state what the Stiefel–Whitney classes are.

  • 2
    $\begingroup$ This doesn't answer your question, but it may be of interest. $\endgroup$ May 13, 2022 at 0:26
  • $\begingroup$ @ Michael Albanese Thanks! $\endgroup$ May 13, 2022 at 1:46

1 Answer 1


There is a fibration $SU(n) \overset p\to SU(n)/SO(n) \overset j\to BSO(n)$, where the $j$ is the classifying map of $p$, viewed as (the projection of) a principal $SO(n)$-bundle. The Stiefel–Whitney classes for your Wu-esque manifolds are the $j^*$-images of the universal Stiefel–Whitney classes; this is in Borel and Hirzebruch's "Characteristic classes and homogeneous spaces I."

To find a nonzero Stiefel–Whitney number is to find a product of these classes of total degree $\dim SU(n)/SO(n)$. The computation in Mimura–Toda shows that the cohomology over $\mathbb F_2$ is an exterior algebra on one generator each of degrees $2$ through $n$, and that these are the images of the universal Stiefel–Whitney classes other than $w_1$. The product of these generators thus does represent the fundamental class.

  • 6
    $\begingroup$ Thanks for the lightning-fast response! When I write you in the acknowledgments, do you have a preferred name over your mathoverflow id? $\endgroup$ May 13, 2022 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.