I am looking for some analogous nontrivial but known statements and references about statements of the form:

Every _______ $d$-manifold has an $S$-structure.

Here _______ is a placeholder for certain conditions, like orientability, complex, triangulable, PL, topological, etc.

This is very vague. What do I really mean?

For example:

  1. Every orientable $4$-manifold has a $\mathit{Spin}^c$ Structure is true.

The precise statement on J. W. Morgan's "The Seiberg-Witten Equations and Applications to the Topology of Smooth Four-Manifolds (MN-44)" that a $4$-manifold $X$ admits a $\mathit{Spin}^c$ structure (Lemma 3.1.2) seems to be that every orientable $4$-manifold $X$ admits a $\mathit{Spin}^c$ structure. This means that we impose the $w_1(X)=0$ for the orientable $4$-manifold $X$. Note that

$$Spin^c=\frac{(Spin \times U(1))}{\mathbb{Z}/2\mathbb{Z}}.$$

  1. Every nonorientable $4$-manifold has a $\mathit{Pin^c}$, $\mathit{Pin^{\tilde c+}}$ or $\mathit{Pin^{\tilde c-}}$ structure is false.

Note that $$\mathit{Pin^c}=\frac{(\mathit{Pin}^+ \times U(1))}{\mathbb{Z}/2\mathbb{Z}}=\frac{(Pin^- \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ $$\mathit{Pin}^{\tilde c+}=\frac{(\mathit{Pin}^+ \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ $$\mathit{Pin}^{\tilde c-}=\frac{(\mathit{Pin}^- \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}}.$$

  1. Spin-H structures

$$\mathit{Spin}^H (n)=\mathit{Spin}(n) \times \mathrm{SU}(2)/{\mathbb{Z}/2\mathbb{Z}}=\mathit{Spin}(n) \times \mathrm{SU}(2)/\{ 1,-1\}$$

Do we have similar statements analogous to every _______ $d$-manifold has a $\mathit{Spin}^H$ structure? Such as every orientable $d$-manifold has a $\mathit{Spin}^H$ structure for some $d$? (like $d=5$?)

  1. Are there other similar statements?
  • 2
    $\begingroup$ All the examples you gave are examples of $G$-structures, so maybe using '$G$-structure' instead of '$S$-structure' would be a more informative title. Unless you're interested in a broader question. $\endgroup$ – Michael Albanese Mar 17 at 22:15
  • $\begingroup$ @MichaelAlbanese isn't your replacement for "..." overly emphatic? I hesitated to change into simply "(...)"; I agree I was puzzled by "..." but I'm skeptical about forged math mode. $\endgroup$ – YCor Mar 17 at 22:32
  • 1
    $\begingroup$ Actually in the title I'd even suggest to even replace it with words, but maybe it's better if the OP does it. The long underscore catches the eye a bit too much in the front list (to my taste). $\endgroup$ – YCor Mar 17 at 22:54
  • 1
    $\begingroup$ @Sebastian: I don't know how Gauss-Bonnet can be used to prove that. More generally, any orientable three-manifold admits a spin structure. Similarly, any closed three-manifold admits a pin$^-$ structure. $\endgroup$ – Michael Albanese Mar 18 at 22:45
  • 1
    $\begingroup$ Use the Euler characteristic: it's even. $\endgroup$ – Tyrone Mar 19 at 9:33

In a paper in preparation, Aleksandar Milivojevic and I show that every orientable five-manifold admits a spin${}^h$ structure, and we provide examples of (higher-dimensional) orientable manifolds which do not admit a spin${}^h$ structure.

The proof that every orientable five-manifold admits a spin${}^h$ structure is relatively simple. First, a spin${}^h$ structure on an orientable $n$-manifold is equivalent to the existence of an immersion into a spin $(n + 3)$-manifold. Now, by Cohen's immersion theorem, every five-manifold immerses into $\mathbb{R}^8$, which is spin.

The construction of orientable manifolds which do not admit a spin${}^h$ structure is harder. We construct examples in dimension $8$ and above. We do not know whether there are such manifolds in dimensions $6$ or $7$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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