User:Silly rabbit/properties as a tensor

Properties as a tensor
For each finite-dimensional vector space, denote by L2(V) the space of all bilinear mappings


 * $$g : V\times V\to \mathbb{R}.$$

By the universal property of the tensor product and the tensor-hom adjunction, there are natural isomorphisms


 * $$L^2(V) \stackrel{\cong}{\to} (V\otimes V)^* \stackrel{\cong}{\to} \operatorname{Hom}(V,V^*)\stackrel{\cong}{\to} V^*\otimes V^*$$

between L2(V), the dual space of the tensor product V &otimes; V, the space of linear mappings from V to its dual, and the tensor product of the dual space of V with itself.

Specifically, if g is a bilinear mapping on V, then the element g&otimes; of (V&otimes;V)* naturally associated to g is defined on simple elements of V&otimes;V by


 * $$g_\otimes(v\otimes w) = g(v,w)$$

extended by linearity to arbitrary linear combinations of elements of V&otimes;V. The original bilinear form g is symmetric if and only if
 * $$g_\otimes\circ\tau = g_\otimes$$

where
 * $$\tau : V\otimes V\stackrel{\cong}{\to} V\otimes V$$

is the braiding map.

For the second, a bilinear form g on V gives rise to a mapping


 * $$S_g : V\to V^*$$

from V into its dual space via


 * $$S_g u = g(u,\cdot)$$

for all u &isin; V. Conversely, a linear mapping S of V to V* gives rise to a bilinear maping gS by


 * $$g_S(u,v) = [Su,v].$$

The bilinear form g is symmetric if and only if the mapping Sg is symmetric in the sense that


 * $$[S_gu,v] = [S_gv,u] $$

for all u, v &isin; V, where [•,•] denotes the pairing of V with its own dual space. Furthermore, g is non-degenerate if and only if Sg is a linear isomorphism.

Finally, given an S &isin; Hom(V,V*), the transpose of S defines an element of Hom(V*,V**) = Hom(V*,V) by the double-dual isomorphism.
 * $$\operatorname{Hom}(V,V^*)\cong V^*\otimes V^*.$$

Taking the vector space V to be the tangent space TpM at a fixed point p &isin; M, a metric g determines a tensor

The correspondence $$\scriptstyle{L^2:V\mapsto L^2(V)}$$ is a contravariant functor on the category of finite-dimensional vector spaces. furthermore, this functor is smooth in the sense that


 * $$L^2 : \operatorname{Hom}(V,W) \to \operatorname{Hom}(L^2(W),L^2(V))$$

is a smooth map of vector spaces. By descent, L2 induces a base-preserving functor on vector bundles over M such that, on the fibres,


 * $$L^2(E)_p = L^2(E_p)$$

for every finite-dimensional vector bundle E and point p in M.

If V is any vector space, then by the universal property of the tensor product, there are natural isomorphisms


 * $$L^2(V) \stackrel{\cong}{\to} (V\times V)^* \stackrel{\cong}{\to} \operatorname{Hom}(V,V^*)$$

between L2(V), the dual space of the tensor product V &otimes; V, and the space of linear mappings from V to its dual.

If V is finite-dimensional, then there is another natural isomorphism