How to identify $\bigwedge^2 TM$ with $\mathfrak{so}(n)$?

Question

Let $(M,g)$ be a Riemannian manifold and $TM$ its tangent bundle. Does anyone know how to identify $\bigwedge^2 TM$ with $\mathfrak{so}(n)$?

Answer 1

First of all, I don't think you can identify $\bigwedge^2TM$ with $\mathfrak{so}(n)$, unless the tangent bundle is trivial.

Point-wise, however, $\bigwedge^2T_xM$ consists of skew-symmetric second-order "contravariant" tensors at $x\in M$. These can be "lowered" by the metric to create "anti-self-adjoint" operators. In index notation, $A=\sum_{i,j}A^{ij}\partial_i\otimes\partial_j$ is in $\bigwedge^2T_xM$, if $A^{ij}=-A^{ji}$.

Then, we can use the metric to do a partial "lowering" as $\flat A=\sum_{i,j}A^i_{\ j}\partial_i\otimes dx^j$, where $A^i_{\ j}=\sum_kA^{ik}g_{kj}$. This tensor $\flat A$ can be seen as a linear transformation on $T_xM$. For any $v\in T_xM$, $$ \flat A(v)=\sum_{i,j}A^i_{\ j}v^j\partial_i. $$

It can be checked that this operator obeys the anti-self-adjointness condition $$ g(u,\flat A(v))=-g(\flat A(u),v). $$ We prove this by calculating in coordinates: $$ g(u,\flat A(v))=\sum_{i,j,k}g_{ij}u^iA^j_{\ k}v^k=\sum_{i,j,k,l} g_{ij}u^iA^{jl}g_{lk}v^k=-\sum_{i,j,k,l}g_{ij}u^iA^{lj}g_{lk}v^k \\ =-\sum_{i,j,k,l}u^iA^{l}_{\ i}g_{lk}v^k=-g(\flat A(u),v).$$

But opeators that obey the anti-self-adjointness condition are precisely the operators that can be identified with $\mathfrak{so}(n)$, because an element of $O(n)$ is defined (on $T_xM$) as a linear transformation that satisfies $g(B(u),B(v))=g(u,v)$, so if $B(t)$ is a smooth family of such operators, and $B(0)=Id$, then $$ \frac{d}{dt}g(B(t)u,B(t)v)|_{t=0}=g(B'(0)u,v)+g(u,B'(0)v)=0, $$ so if $A=B'(0)$, then $$ g(Au,v)=-g(u,Av). $$

This is for a single tangent space.

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For a relation on bundles, we can construct a vector bundle whose fibers are $\mathfrak{so}(n)$s. It's total space is clearly given by $$ \bigsqcup_{x\in M}\mathfrak{so}(T_xM), $$ but can also be constructed differently:

We let $O(M)$ be the orthonormal frame bundle, which is a principal fiber bundle with structure group $G=O(n)$. $O(n)$ has a representation on $\mathfrak{so}(n)$ given by the adjoint representation $\text{Ad}:O(n)\mapsto GL(\mathfrak{so}(n))$, so we can construct an associated vector bundle as follows - we consider the set $O(M)\times\mathfrak{so}(n)$, with the equivalence relation $(e,A)\sim(e\Lambda,\text{Ad}(\Lambda^{-1})A)$ ($e\in O(M)$ and $\Lambda\in O(n)$). We call the quotient $$ O(M)\times_{O(n)}\mathfrak{so}(n)=O(M)\times\mathfrak{so}(n)/\sim, $$ and can be naturally equipped with a vector bundle structure, whose model fiber is $\mathfrak{so}(n)$.

It is this vector bundle which is isomorphic with $\bigwedge^2TM$ via the fiberwise vector space isomorphism detailed in the beginning of this answer.