Dual element for an inner product

Question

Let $V$ be a finite dimensional real vector space, and $g(\cdot,\cdot)$ a positive definite symmetric bilinear form. For a choice of orthonormal basis $\{e_i\}_i$ with respect to $g$, then it clear that the element $$ g^* := \sum_i g(e_i,e_i) e^i \otimes e^i \in V^* \otimes V^* $$ is well-defined irrespective of the choice of basis (here we have of course denoted the dual basis of $\{e_i\}_i$ by $\{e^i\}_i$). Now using the musical isomorphisms we have a corresponding element $g_*$ in $V \otimes V$. Is it clear that this element will always depend on the choice of basis?

Answer 1

Let $\{e_i\}$ denote an orthonormal basis of $V$. The bilinear form $g\in Hom(V\otimes V,\mathbb K)$ can be identified with the element $g^{*}\in V^{*}\otimes V^{*}$ using the isomorphism

$$Hom(V\otimes V,\mathbb K):=(V\otimes V)^{*}\rightarrow V^{*}\otimes V^{*}. $$

Explicitly

$$g^{*}=g(e_i,e_j)e^i\otimes e^j$$

using the Einstein summation convention and denoting by $\{e^i\}$ the dual basis of $\{e_i\}$ in $V^{*}$.

Under the isomorphism $V\rightarrow V^{*}$ induced by the above choice of basis, we can define the element $g_{*}\in V\otimes V$ via

$$g_{*}=g(e_i,e_j)e_i\otimes e_j.$$

If $\bar{e}_i:=a_{ij}e_j$ is a new orthonrmal basis for $V$, with $a^{-1}=a^t$, under the basis transformation $e_i\mapsto \bar{e}_i$ the tensor $g_{*}$ transforms as follows:

$$g(\bar{e}_i,\bar{e}_j)\bar{e}_i\otimes\bar{e}_j= a_{ik}a_{jl}a_{ir}a_{js}g(e_k,e_l)e_r\otimes e_s;$$

using orthonormality of $a$ (i.e. $a_{ik}a_{ir}=a^{t}_{ki}a_{ir}=a^{-1}_{ki}a_{ir}=\delta_{kr}$ and similarly for $a_{jl}a_{js}$ ) we arrive at

$$g(\bar{e}_i,\bar{e}_j)\bar{e}_i\otimes\bar{e}_j= \delta_{kr}\delta_{ls}g(e_k,e_l)e_r\otimes e_s=g(e_r,e_s)e_r\otimes e_s=g(e_i,e_j)e_i\otimes e_j,$$

as the name of repeated indices is not relevant. In other words, $g_{*}$ is invariant under the orthonormal basis transformation $e_i\mapsto \bar{e}_i$.