Editing Metric tensor (section)

===Metric as a section of a bundle===
By the [[Tensor product#Universal property|universal property of the tensor product]], any bilinear mapping ({{EquationNote|10}}) gives rise [[natural transformation|naturally]] to a [[section (fiber bundle)|section]] {{math|''g''<sub>⊗</sub>}} of the [[dual space|dual]] of the [[tensor product bundle]] of {{math|T''M''}} with itself

:<math>g_\otimes \in \Gamma\left((\mathrm{T}M \otimes \mathrm{T}M)^*\right).</math>

The section {{math|''g''<sub>⊗</sub>}} is defined on simple elements of {{math|T''M'' ⊗ T''M''}} by

:<math>g_\otimes(v \otimes w) = g(v, w)</math>

and is defined on arbitrary elements of {{math|T''M'' ⊗ T''M''}} by extending linearly to linear combinations of simple elements. The original bilinear form {{mvar|g}} is symmetric if and only if
:<math>g_\otimes \circ \tau = g_\otimes</math>
where
:<math>\tau : \mathrm{T}M \otimes \mathrm{T}M \stackrel{\cong}{\to} TM \otimes TM</math>
is the [[tensor product#Tensor powers and braiding|braiding map]].

Since {{mvar|M}} is finite-dimensional, there is a [[natural isomorphism]]

:<math>(\mathrm{T}M \otimes \mathrm{T}M)^* \cong \mathrm{T}^*M \otimes \mathrm{T}^*M,</math>

so that {{math|''g''<sub>⊗</sub>}} is regarded also as a section of the bundle {{math|T*''M'' ⊗ T*''M''}} of the [[cotangent bundle]] {{math|T*''M''}} with itself. Since {{mvar|g}} is symmetric as a bilinear mapping, it follows that {{math|''g''<sub>⊗</sub>}} is a [[symmetric tensor]].