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Forgive me if I mess some of these concepts up or say something incorrect, I am still figuring out all the details of an Ehresmann connection in an associated vector bundle.

So, how do we relate these two practically? Like for a two sphere I can very easily write down the levi civita connection in the standard way for the usual round metric, but if I wanted to get something equivalent by starting with a connection in the frame bundle I don't really know where to begin.

Specifically, it seems we would have a natural choice of a horizontal distribution on the frame bundle if we have imposed a metric on it (i.e. the horizontal vector fields are those such that $g(\mathfrak{gl},H)=0$, for a metric $g$), but I'm not sure what other conditions I would need to get the equivalent of the levi civita connection in the tangent bundle.

Or would we want to look at an orthonormal frame bundle? I don't see why we would necessitate that other than that maybe the structure group is smaller as it would be $O(n)$ instead of $GL(n)$, but at that point why not consider the bundle of orthonormal frames with the same orientation and use $SO(n)$?

I think I am missing something

Chris
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  • https://math.stackexchange.com/questions/1407800/how-do-connection-1-form-and-ehresmann-version-of-connections-relate-to-each-oth – infinitylord May 07 '22 at 22:21
  • @infinitylord I don't see how this answers my question... – Chris May 07 '22 at 23:30
  • I'm not sure I understand what your specific question is? Using a horizontal distribution you can define a $\mathfrak g$-valued 1-form $A$ on $P$. Then you can define a covariant derivative operator using horizontal lifts of vectors from $M$ to $P$, or using a formula like $\nabla_A = \mathrm d + A$. You can go the other way and define horizontal subspaces in $TP$ to be the span of parallel transported frames by a covariant derivative $\nabla_A$. – Keshav May 08 '22 at 03:40
  • What do you mean by an "associated vector bundle"? – Deane May 08 '22 at 04:20
  • @Keshav I’m asking what extra conditions do I need to place on the horizontal sub bundle of the total space to obtain something equivalent to the Levi civita connection in the tangent bundle. – Chris May 08 '22 at 13:37
  • @Deane given a principal bundle $P$ with structure group $G$, a representation $\rho$, and vector space $V$, the associated vector bundle is the quotient space $(PxV)/G$, with the right action given by: $(pg, \rho(g^{-1})v)$. – Chris May 08 '22 at 14:01
  • The two conditions for the Levi-Civita connection are that it is metric-compatible and torsion-free. These can be translated over to the principal (frame) bundle world. Putting a metric on $TM$ is equivalent to reducing the structure group of $P$ to $O(n)$, and metric compatibility comes from working with (connections on) this new bundle. The torsion tensor I believe can be written down in terms of principal bundle language, but I don't know it right now. That would give you another equation / condition. – Keshav May 08 '22 at 21:09
  • @Keshav Why is putting a metric on $TM$ equivalent to reducing the structure group of $P$ to $O(n)$? Like I see with a metric we can reduce to the structure group to $O(n)$, but I don't see why those are equivalent. – Chris May 08 '22 at 23:27
  • @ChristopherQuinnLaFondJr. One potential cause for confusion here is that we can't treat the frame bundle as an abstract principal bundle, we need to encode the link between $FM$ and $TM$ e,g, by using a solder form. This allows one to define torsion and make the connection between metrics on $TM$ and $O(n)$ structures on $FM$ precise. – Kajelad May 08 '22 at 23:46
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    @ChristopherQuinnLaFondJr. If $P$ is the frame bundle, (i.e. each point in $P$ is a frame of $T_xM$ at a point $x$ in $M$), and then you reduce to a sub $O(n)$ bundle inside of $P$, then this specifies (at each point $x$) which frames can be called "orthogonal". This is sufficient to defining the metric at $T_xM$ (It looks like the identity matrix in "this frame"). – Keshav May 09 '22 at 00:34
  • @Kajelad, could you elaborate a little on how a solder form allows us to do that? the wiki article you linked wasnt very illuminating for me Im afraid. – Chris May 09 '22 at 22:06
  • @ChristopherQuinnLaFondJr. A solder form on $P$ gives you an isomorphism between $TM$ and the associated bundle $P\times_GV$ where $V$ is the obvious choice of rep. a metric on $TM$ then gives the $GL(n)$ to $O(n)$ reduction. Also, the solder form can be used to define the torsion of a principal connection directly, which corresponds with the torsion of the associated affine connection. – Kajelad May 10 '22 at 01:51

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For those who come across this question, an explicit answer is handled in depth in the Kobayashi’s foundations of differential geometry. In the special case where $E$ is the tangent bundle, one can think of it as a vector bundle associated to the bundle of orthonormal frames of $TM$. One can then construct a notion of torsion in the frame bundle, and we can find a torsion free, metric compatible connection one form in the bundle of orthonormal that corresponds to the Levi-Civita connection.

Chris
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  • Sorry to comment on an older post but could you elaborate more on this? For a manifold $M$, we define the Levi-Civita connection $\nabla$ on $TM$. Once we fix a pseudo-Riemannian metric we get an isomorphism between $TM$ and an associated orthogonal frame bundle. But how does $\nabla$ does define a Lie algebra valued connection on the associated orthogonal frame bundle? – CBBAM Jan 07 '24 at 03:52
  • @CBBAM First note that in any local orthonormal frame $e_i$, $\nabla$ can be viewed as a collection of one forms $\xi^i_j$ which satisfy $\nabla e_i=\xi^j_i\otimes e_j$, and convince yourself from the metric compatibility conditions that the components $\xi^i_j$ define a one form on an open set $U$ with values in $\mathfrak{so}$. Now using $\xi^i_j$, define Lie algebra valued one forms on $U\times SO(n)$ which satisfy the definition of a connection, and use the local trivializations to pull them back on principal bundle $SO(M)$. Check they agree on overlaps and you have a global conection. – Chris Jan 08 '24 at 01:07
  • Thank you for the response! I am able to see that $\nabla$ locally defines a matrix of 1-forms. But I am having trouble seeing how the metric compatibility implies that they must be $\mathfrak{so}$ valued. Wouldn't we also have to worry about the torsion free condition somehow being satisfied? – CBBAM Jan 08 '24 at 03:22
  • @CBBAM No, any metric compatible connection will induce a matrix of $\mathfrak{so}$ valued one forms. Write the metric compatibility condition in a specific orthonormal frame and you will see. The torsion free criteria just makes the connection the unique levi civita connection. – Chris Jan 08 '24 at 06:38
  • I have been trying to work out the details based on your last comment but I'm still a little unclear on everything. Suppose $\nabla$ is a connection compatible with the metric $g$ on the associated $SO(M)$ vector bundle. Then for any $X \in \mathfrak{X}(M)$ and sections $s, t$ of the associated bundle we have $$Xg(s,t) = g(\nabla_X s, t) + g(s, \nabla_X t).$$ – CBBAM Jan 08 '24 at 19:13
  • I have three main points of confusion. (1) How does the equation in my last comment imply that the induced 1-forms must be $\mathfrak{so}$ valued? (2) What does this have to do with the Levi-Civita connection on $TM$? (3) How does all this work on the associated vector bundle and $TM$ tie back to the principal $SO(M)$ frame bundle? – CBBAM Jan 08 '24 at 19:15
  • Let $s=e_i$ and $t=e_j$, then $Xg(e_i,e_j)=0$, and we have have $g(\nabla_Xe_i,e_j)= g(\xi(X)i^ke_k,e_j)=\xi(X)^k_i\delta{kj}$. Put this all together to find that for all vector fields $\xi(X)^j_i+\xi(X)^i_j=0$. Since this holds for all vector fields, it follows $\xi^j_i+\xi^i_j=0$ for all $i$ and $j$, so $\xi$ is a one form with values in $\mathfrak{so}$. I'll leave it to you to handle the pseudo riemannian case, you just have to be careful with the signature. – Chris Jan 08 '24 at 20:01
  • @CBBAM also, the vector bundle associated to $SO(M)$ via the standard representation $\textit{is}$ the tangent bundle. We are a starting with a metric compatible connection on $TM$, and using it to construct a connection one form on $SO(M)$. This will in turn yield a connection on $TM$ via the standard theory that is equal to the levi-civita connection. – Chris Jan 08 '24 at 20:04
  • This is wonderful, thank you very much! So we start with $\nabla$ which we always choose to be the Levi-Civita connection on $TM$, and then from the argument you gave we get a $\mathfrak{so}$ valued 1-form $\xi$ which can be shown to be a connection on the principal $SO(M)$ frame bundle? – CBBAM Jan 08 '24 at 20:38
  • The only two parts that are still a little confusing to me are (1) why we even bother with the associated vector bundle at all? Even though it is equivalent to $TM$, it seems like we can carry out this whole construction using only $TM$ and $SO(M)$. (2) Can you elaborate more on your last sentence? That is, if we assume $\nabla$ is just a metric connection (not necessarily Levi-Civita on $TM$) then how does this construction imply that it must be the Levi-Civita connection? – CBBAM Jan 08 '24 at 20:38
  • @CBBAM It is at times easier to talk about connections on a principal bundle, and induced connections on an associated vector bundle than it is to talk about connection on a vector bundle in generality. It turns out these two points of view are equivalent, so you can always translate between the two, the translation is useful. If we assume $\nabla$ is just a metric connection, then we get a unique (in the sense that the metric connections on $TM$ are in bijection with connection one forms on $SO(M)$) connection one form on $SO(M)$, but this does not have to be the Levi-Civita connection. – Chris Jan 08 '24 at 20:51