# Writing a tangent bundle in terms of the rotation group of the embedding space and connection (parallel transport)

Ok, so I'm a physics student, but this is primarily a math question, so I'm posting it here. I apologize in advance for any misuse of terminology and formality. Looking at the Tangent bundle for an m-dimensional (possibly pseudo-) Riemannian manifold $$M$$, we write it as the disjoint union of the tangent space at each point in the manifold.

$$TM=\dot{\bigsqcup}_{x\in M}T_{x}M$$

I can't help but wonder: is there another way to write this? Let us consider the manifold as embedded in $$\mathbb{R}^{n}$$. Then the tangent space at a point x $$\left(T_{x}M\right)$$ is related to the tangent space at another point q by a rotation in $$\mathbb{R}^{n}$$. Denoting such a point dependent transformation $$A_{pq}$$ it is obvious that $$A_{pq}\in O(n)$$ (or $$O(n,k)$$ for a pseudo-Riemannian manifold). It is clear then that our tangent bundle can be described simply by:

$$TM=\left(\dot{\bigsqcup}_{x\in M}A_{xo}\right)T_{o}M$$

Where the point o is an arbitrary fixed point on M. This is simply: $$TM=\left(\dot{\bigsqcup}_{x\in M}A_{xo}\right)\mathbb{R}^{m}$$

Which is to say we can write it in terms of the union of the tangent space at one point with all of the position dependent (a function of the embedded manifold in question) transformations which are elements of the aformentioned rotation group.

Is this correct so far? Now when we write the projection $$\pi$$ to $$M$$, how exactly do we go about doing this? Do we get a connection in the covariant derivative with values in projective representations of the rotation group on M?

For example, a Lorentzian manifold, locally has $$O(3,1)$$ as the rotation group, can we then say these Lorentz valued connections are projective representations from $$O(n,k)$$? I believe these are then spinor representations? Thank you very much in advance!

For exemplify what I'm saying, If M is flat $$\mathbb{R}^{3,1}$$ then the embedding and projections are trivial, but If I have say an $$S^{3}xS^{1}$$ space then I would have non-vanishing elements of the $$O(4,2)$$ appearing in a connection? Thank you very much in advance! I asked a similar question here but on the physics stackexchange they seem to be missing the point of my question.

• Note that your $A_{x0}$ might not be continuous (with respect to $x$). Commented Feb 9, 2021 at 22:57
• @ArcticChar Not sure I follow you there please explain? Commented Feb 10, 2021 at 9:07

"Then the tangent space at a point $$x$$ ($$𝑇_𝑥 𝑀$$) is related to the tangent space at another point $$q$$ by a rotation in $$\Bbb R^n$$."
Consider the case of a circle in the $$xyzw$$-plane of $$\Bbb R^4$$. The tangent space at $$(1, 0, 0,0)$$ consists of multiples of $$(0,1, 0,0)$$; the tangent space at $$(0, 1,0,0)$$ consists of mutliples of $$(-1, 0, 0,0)$$. And rotation by 90 degrees in the $$xy$$-plane takes one of these to the other. But the rotation defined by left-multiplication by the matrix $$A_u = \pmatrix{ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & c & -s \\ 0 & 0 & s & c}$$ where $$s = \sin u, c = \cos u$$ ... that rotation also takes one tangent space to the other. In particular, the $$A$$ in your description is not unique, and it's not even clear that there's a continuously-varying function $$A : M \times M \to SO(n)$$ such that $$A(p,q)$$ sends $$T_p M$$ to $$T_q M$$. (There surely is in this example of mine, but when you think about a Mobius strip....things get a bit dicier.
• +1 Thank you! I think the ambiguity in A in your example comes from adding extra dimensions instead of embedding in $\mathbb{R}^{2}$ though I definitely see your point with the Mobius strip! Even in that case however, can I not relate the tangent spaces at two "nearby" points via a rotation in the embedding space, such that a covariant derivative has this structure group for it's connection? I'm really just trying to work out these structure groups for different topologies when written as an embedding. Commented Feb 9, 2021 at 23:17