Product connection on tensor and hom bundles

Give each of finitely many bundles $E_j$ a connection. Use the tensor bundle constructed in Finite tensor products of smooth vector bundles, including its multilinear model on the dual fibres. The product connection is characterized on local elementary tensors by ${\mathrm{\nabla }}_X(s_1\otimes \cdots \otimes s_k)={\sum }_j{s}_1\otimes \cdots \otimes {\mathrm{\nabla }}_X^j{s}_j\otimes \cdots \otimes s_k\mathrm{.}$ For the empty product, the connection on scalar functions is ${\mathrm{\nabla }}_{X}f=X(f)$.

To construct it without a decomposition assumption, for a multilinear-model section $T$ and local dual sections ${\alpha }_j$ put $({\mathrm{\nabla }}_{X}T)({\alpha }_1,\dots ,{\alpha }_k)=X(T({\alpha }_1,\dots ,{\alpha }_k))-{\sum }_{j}T({\alpha }_1,\dots ,({\mathrm{\nabla }}^j{)}_X^{\ast }{\alpha }_j,\dots ,{\alpha }_k)\mathrm{.}$ Use the dual connections of Dual connection. Multiplying any argument by a smooth $f$ produces two opposite $X(f)T({\alpha }_1,\dots ,{\alpha }_k)$ terms; thus this expression is function-linear in every dual argument. In dual frames its coefficients are smooth and depend only on their fibre values, so it defines a smooth tensor section. The same intrinsic expression on overlaps gives agreement. Its real-linearity, function-linearity in $X$, and rule ${\mathrm{\nabla }}_X(fT)=X(f)T+f{\mathrm{\nabla }}_{X}T$ follow by direct expansion, establishing the connection. This also gives the gluing criterion of Local connection forms glue exactly when they obey the transformation law.

On a pure tensor, the scalar product rule and the dual pairing identity reduce this expression to the first displayed sum. Hence it respects the balanced identity $(fs)\otimes t=s\otimes (ft)$: either derivative gives the same term $X(f)s\otimes t$ and the same two terms multiplied by $f$. Every section has a finite local product-frame expansion, so that rule and the connection Leibniz identity force uniqueness.

For bundles $E,F$, identify $F\otimes E^{\ast }$ with $\mathrm{Hom}(E,F)$ by $t\otimes \alpha \mapsto (v\mapsto \alpha (v)t)$. In local bases these elementary tensors are precisely the matrix units, proving this is a smooth fibrewise isomorphism. The induced Hom connection is therefore $({\mathrm{\nabla }}_{X}A)(s)={\mathrm{\nabla }}_X^F(A(s))-A({\mathrm{\nabla }}_X^{E}s)\mathrm{.}$ The two $X(f)A(s)$ terms cancel when $s$ is replaced by $fs$, ensuring a fibrewise map. Rank-zero factors give zero bundles, whereas the empty product is the scalar line; these conventions are different. No choice beyond finitely many local frames is needed.