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How do the oriented orthonormal frame bundles of product spaces $M=M_{1}\times M_{2}$ work?

For example, I'm interested in the manifold $M=M_{1}\times M_{2}=S^{3}\times S^{1}$. I know that the oriented orthonormal frame bundle of each one $OF(M_{i})$ is:

$$OF(S^{3})=SO(4)$$

where:

$$S^{3}=\frac{SO(4)}{SO(3)}$$

And

$$OF(S^{1})=SO(2)$$

Note, that I'm working with a pseudoriemannian metric such that $S^{1}$ is associated with negative signature in the quadratic form.

I'm guessing, in analogy with Minkowski space $R^{3,1}$ that we have:

$$OF(S^{3}\times S^{1})=SO(4,2)$$

Is it then going to be true that:

$$SU(2)\times U(1)=\frac{SO(4,2)}{SO(3,1)}$$

Where we have used the isomorphisms $SU(2)=S^{3}$ and $S^{1}=U(1)$.

Or, put another way, can we then write the Principle oriented orthonormal frame bundle as:

$$OF(S^{3}\times S^{1})=SU(2)\times U(1)\times SO(3,1)$$

However clearly:

$$SO(4,2)\neq SU(2)\times U(1)\times SO(3,1)$$

??? can someone explain what is the correct way? How would I map this (the OF(M) to the tangent bundle?

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  • $\begingroup$ Do you want to explicitly identify the total space or just understand what the bundle is? Often times authors just define the bundle locally via transition functions rather than describing the total space. $\endgroup$
    – pancini
    Sep 22, 2020 at 0:15
  • $\begingroup$ @ElliotG I was hoping to understand both starting with the total space, but I"ll take whatever you offer up and be thankful. $\endgroup$
    – R. Rankin
    Sep 22, 2020 at 0:17
  • $\begingroup$ Your "analogy" doesn't apply in this case. The frame bundle of a product space is, generally speaking, much more complicated than the frame bundle of its factors, and there is no reason to expect it to be isomorphic to a quotient of Lie groups. It might be more fruitful to start with the isometric embedding into $\mathbb{R}^{4,2}$. $\endgroup$
    – Kajelad
    Sep 26, 2020 at 19:53
  • $\begingroup$ @Kajelad Thank you, I'm working on the embedding right now. I can say with certainty that the principle bundle of oriented orthonormal frames is trivial on the manifold (there exists a global section) due to the nature of the factors. It seems like that should simplify things a bit for me? $\endgroup$
    – R. Rankin
    Sep 27, 2020 at 18:29
  • $\begingroup$ By "the manifold" do you mean $S^3\times S^1$ or $\mathbb{R}^{4,2}$? There's no need to use the frame bundle $F_{SO}(\mathbb{R}^{4,2})$; the embedding allows you to write down the elements of $F_{SO}(S^3\times S^1)$ as collections of vectors subject to constraints. If you haven't done something like this already, it might be a good idea to start with a simpler case like describing $F_{O}(S^2)$ using the embedding into $\mathbb{R}^{3}$. $\endgroup$
    – Kajelad
    Sep 27, 2020 at 18:57

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