When we study the frame bundle of a n-dimensional Riemannian manifold $M$ we start with a principal $GL(n,\mathbb{R})$ bundle over $M$. There are a series of topological obstructions to reducing the structure. If the bundle is orientable we have a reduction $$GL^{+}(n,\mathbb{R})\rightarrow GL(n,\mathbb{R}) $$

If the manifold admits a metric (always since we're talking Riemannian), then we get:

$$O(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$$

We call it a reduction to the orthonormal frame bundle. Suppose we look at instead:

$$SL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$$

Then we have a volume form on our manifold. I'm familiar with the idea, what I'm wondering about is a reduction of the form:

$$SL(n,\mathbb{C})\rightarrow SL(2n,\mathbb{R})$$

for a manifold of even dimension. What does that correspond to? What are the topological obstructions to it's existence? Or we might consider instead:

$$GL(n,\mathbb{C})\rightarrow GL(2n,\mathbb{R})$$

is this just the existence of a complex structure, or is there more, what kind of frames does this reduction of the frame bundle correspond to?


1 Answer 1


Let me answer the $GL$ version of your question (from the last paragraph). Suppose that $P\to M$ is a principal $GL(2n, {\mathbb R})$-bundle over a manifold $M$ (this need not be the frame bundle of the tangent bundle). Let $E\to M$ be the associated ${\mathbb R}^{2n}$-bundle.

I am assuming that $GL(n, {\mathbb C})$ is embedded in $GL(2n, {\mathbb R})$ in the standard fashion: $$ A+iB\mapsto A\oplus B $$ (the image is a block-diagonal matrix). Then each reduction of $P$ to a $GL(n, {\mathbb C})$-bundle just amounts to a choice of a complex structure $J$ on the bundle $E\to M$ (and vice-versa). In the case when $E\to M$ is the tangent bundle over $M$, the complex structure $J$ on $E\to M$ is called an almost complex structure on $M$. (To be a genuine complex structure, it has to satisfy an integrability condition.) In terms of frames, this can be described as follows: Suppose that $J$ is a complex structure on the bundle $E\to M$ and for $x\in M$, $(v_1,...,v_n)$ is a basis in the complex vector space $(E_x,J)$ (the fiber over $x$). Then the corresponding real frame (a real basis in $E_x$) will be $$ (v_1,...,v_n, Jv_1,..., Jv_n). $$ Conversely, a real frame $$ (v_1,...,v_n, w_1,...,w_n) $$ comes from a complex frame if and only if $w_k=Jv_k$, $k=1,...,n$.

If you really want to understand the $SL$-case, you will have to deal with a complex volume form $\omega\in \Omega^{n,0}(M)$ (nondegenerate at every point $x\in X$). Then complex frames $(v_1,...,v_n)$ will be required to satisfy $$ \omega(v_1,...,v_n)=1. $$ Since you did not explain a motivation for this, I will stop here.

  • $\begingroup$ Excellent! Thank you very much! answers my question completely. $\endgroup$
    – R. Rankin
    Dec 26, 2022 at 18:26
  • $\begingroup$ Ps. Motivation for the latter case was an interest in when we get deformation retractions even of $SL(n,\mathbb{C})$ to $SU(n)$. I'm just trying to understand the whole sequence of reductions starting with the general linear group. I'm guessing path taken doesn't matter? Say $GL(n,C)$ to $SL(n,C)$ instead of the other route? $\endgroup$
    – R. Rankin
    Dec 26, 2022 at 19:29
  • $\begingroup$ @R.Rankin: yes, it does not matter which path of reductions you take. $\endgroup$ Dec 26, 2022 at 19:38
  • $\begingroup$ Random thought, Does an $SL(2,\mathbb{C})$ structure on spacetime then correspond to a unit volume form? That would be interesting since we never get told that when we learn about the spinor bundle. This is usually taken via the universal cover of the connected component of $SO(1,3)$. $\endgroup$
    – R. Rankin
    Dec 27, 2022 at 10:01
  • $\begingroup$ @R.Rankin we have to be a little more careful here. In that identification we are thinking of $\operatorname{SL}(2,\mathbb{C})$ as a real form of $\operatorname{SL}(2,\mathbb{C})\times \operatorname{SL}(2,\mathbb{C})$ a reduction to this complexifed version would be a volume form on each (complex) half-spinor bundle. The complexification of $\mathbb{R}^{3,1}$ is then the tensor product of the two half-spinor bundles and a reduction to the real form gives a real structure which identifies $\mathbb{R}^{3,1}$. $\endgroup$
    – Callum
    Dec 28, 2022 at 12:38

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