Tensor product of connections

Definition

Suppose $E,E'$ are vector bundles over a differential manifold $M$. Suppose $\nabla$ is a connection on $E$ and $\nabla '$ is a connection on $E'$. The tensor product $\nabla \otimes \nabla '$ is defined as the unique connection on $E\otimes E'$ such that the following is satisfied for all sections $s,s'$ of $E,E'$ respectively:

$(\nabla \otimes \nabla ')_{X}(s\otimes s')=\nabla _{X}(s)\otimes s'+s\otimes \nabla '_{X}(s')$.

Facts

Well-definedness

Further information: Tensor product of connections is well-defined

It is not completely clear from the definition that the tensor product of connections is well-defined. What needs to be shown is that the definition given above for pure tensor products of sections can be extended to all sections consistently, while maintaining the property of being a connection.

Associativity

Further information: Tensor product of connections is associative upto natural isomorphism

Suppose $E,E',E''$ are vector bundles over a differential manifold $M$, with connections $\nabla ,\nabla ',\nabla ''$ respectively. Then, under the natural isomorphism:

$(E\otimes E')\otimes E''\to E\otimes (E'\otimes E'')$,

the connections $(\nabla \otimes \nabla ')\otimes \nabla ''$ and $\nabla \otimes (\nabla '\otimes \nabla '')$ get identified.

Commutativity

Further information: Tensor product of connections is commutative upto natural isomorphism

Suppose $E,E'$ are vector bundles over a differential manifold $M$, with connections $\nabla ,\nabla '$ respectively. Then, under the natural isomorphism:

$E\otimes E'\to E'\otimes E$

the connections $\nabla \otimes \nabla '$ and $\nabla '\otimes \nabla$ get identified.

Distributivity with direct sum

Further information: Distributivity relation between direct sum and tensor product of connections

Suppose $E,E',E''$ are vector bundles over a differential manifold $M$, with connections $\nabla ,\nabla ',\nabla ''$ respectively. Then, under the natural isomorphism:

$E\otimes (E'\oplus E'')\to (E\otimes E')\oplus (E\otimes E'')$

we have an identification between $\nabla \otimes (\nabla '\oplus \nabla '')$ and $(\nabla \otimes \nabla ')\oplus \nabla \otimes \nabla ''$. Here, $\oplus$ is the direct sum of connections.

An analogous distributivity law identifies $(\nabla \oplus \nabla ')\otimes \nabla ''$ and $(\nabla \otimes \nabla '')\oplus (\nabla '\otimes \nabla '')$.

Commutes with dual connection operation

Further information: Dual connection to tensor product equals tensor product of dual connections