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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.

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

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"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.

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  • $\begingroup$ +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. $\endgroup$
    – R. Rankin
    Commented Feb 9, 2021 at 23:17
  • $\begingroup$ Those dimensions are "extra" in this case only because the circle is particularly simple. If you have a projective plane, for instance, it cannot be embedded in either 3- or 4-space, so 3 "extra" dimensions are absolutely needed, and that's enough to make this same kind of example work. I have no idea what "a covariant derivative has this structure group as its connection" means, so I cannot answer your question. But my example should be regarded as typical rather than exceptional. $\endgroup$ Commented Feb 9, 2021 at 23:23
  • $\begingroup$ I'm really interested in using this in the context of parallel transport. In physics we use the tangent bundle to compare vectors at one point with vectors at another point via parallel transport. In this sense YES there will be a different transformation A for every different path taken between the points, as a vector will be parallel transported differently along different paths, this is actually esential in order to describe curvature. The Connection and covariant derivative determine how vectors change as we move from one tangent space to the next (horizontally through the tangent bundle) $\endgroup$
    – R. Rankin
    Commented Feb 10, 2021 at 9:03

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