# Tensor product of connections

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## 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 Failed to parse (unknown function "\optimes"): (\nabla \otimes \nabla'') \oplus (\nabla' \optimes \nabla'') .