# Representing conversion of (1,1) tensor to (2,0) tensor

• I

## Main Question or Discussion Point

A non-degenerate Hermitian form ##(.|.)## on a vector space ##V## can be identified with a map ##L:V \to V^*## such that ##L(v)=\tilde{v}## and ##\tilde{v}(w) \equiv (v~|~w)##.

Suppose we want to convert a vector ##v## to a dual vector ##\tilde{v}##. In terms of matrices, we can just construct the matrix ##[L]## corresponding to the Hermitian form, and hence the map ##L##, by letting ##L_{ij} = (e_i~|~e_j)##. So

$$\tilde{v}_j = \tilde{v}(e_j) = (v~|~e_j) = \sum_iv^i(e_i~|~e_j) = \sum_i L_{ij}v^i$$

If ##[v]## and ##[\tilde{v}]## are column vectors containing components of ##v## and ##\tilde{v}##, then ##[\tilde{v}] = [L]^T[v]##.

Now I'm trying to apply this whole treatment to the conversion of a ##(1,1)## tensor to a ##(2,0)## tensor. In component representation, the former can be written as ##T_i^{~~j}## and the latter as ##T_{ij}##. From the book I'm reading:

> If we have a non-degenerate bilinear form on ##V##, then we may change the type of ##T## by precomposing with the map ##L## or ##L^{-1}##. If ##T## is of type ##(1,1)## with components ##T_i^{~~j}##, for instance, then we may turn it into a tensor ##\tilde{T}## of type ##(2,0)## by defining ##\tilde{T}(v,w) = T(v,L(w))##.

Given the basis ##\{e_i\}## of ##V##, we have two choices of bases in the dual space: ##\{e^i\}## where ##e^i(e_j) = \delta^i_j##, or ##\{L(e_i)\}## - the latter being the metric dual basis that depends on the choice of the non-degenerate Hermitian form. What is the appropriate choice of basis in this case? I need to confirm this because the matrix representations of ##T## and ##\tilde{T}## would depend on it.

How do I come up with a matrix representation of the conversion from ##T## to ##\tilde{T}##, as was done in the above example? ##\tilde{T}(v,w) = T(v,L(w)) \implies T(v,w) = \tilde{T}(v,L^{-1}(w))##. Given that we've decided on the dual basis, then

$$\tilde{T}_{ij} = \tilde{T}(e_i,e_j) = T(e_i,L(e_j))$$

$$T_i^{~~j} = T(e_i,e^j) = \tilde{T}(e_i,L^{-1}(e^j))$$

I'm assuming that I'll have to express ##L(e_j)## as a linear combination of dual basis vectors ##e^k##'s, and ##L^{-1}(e^j)## as a linear combination of basis vectors ##e_k##'s, but I'm at a loss on how to do that. That's primarily because in the example I gave above, I was able to express vector/covector components in terms of the other's components, but there's no indication on how to do that with vectors/covectors themselves. Any help would be appreciated.

Related Linear and Abstract Algebra News on Phys.org
WWGD
Gold Member
2019 Award
A non-degenerate Hermitian form ##(.|.)## on a vector space ##V## can be identified with a map ##L:V \to V^*## such that ##L(v)=\tilde{v}## and ##\tilde{v}(w) \equiv (v~|~w)##.

Suppose we want to convert a vector ##v## to a dual vector ##\tilde{v}##. In terms of matrices, we can just construct the matrix ##[L]## corresponding to the Hermitian form, and hence the map ##L##, by letting ##L_{ij} = (e_i~|~e_j)##. So

$$\tilde{v}_j = \tilde{v}(e_j) = (v~|~e_j) = \sum_iv^i(e_i~|~e_j) = \sum_i L_{ij}v^i$$

If ##[v]## and ##[\tilde{v}]## are column vectors containing components of ##v## and ##\tilde{v}##, then ##[\tilde{v}] = [L]^T[v]##.

Now I'm trying to apply this whole treatment to the conversion of a ##(1,1)## tensor to a ##(2,0)## tensor. In component representation, the former can be written as ##T_i^{~~j}## and the latter as ##T_{ij}##. From the book I'm reading:

> If we have a non-degenerate bilinear form on ##V##, then we may change the type of ##T## by precomposing with the map ##L## or ##L^{-1}##. If ##T## is of type ##(1,1)## with components ##T_i^{~~j}##, for instance, then we may turn it into a tensor ##\tilde{T}## of type ##(2,0)## by defining ##\tilde{T}(v,w) = T(v,L(w))##.

Given the basis ##\{e_i\}## of ##V##, we have two choices of bases in the dual space: ##\{e^i\}## where ##e^i(e_j) = \delta^i_j##, or ##\{L(e_i)\}## - the latter being the metric dual basis that depends on the choice of the non-degenerate Hermitian form. What is the appropriate choice of basis in this case? I need to confirm this because the matrix representations of ##T## and ##\tilde{T}## would depend on it.

How do I come up with a matrix representation of the conversion from ##T## to ##\tilde{T}##, as was done in the above example? ##\tilde{T}(v,w) = T(v,L(w)) \implies T(v,w) = \tilde{T}(v,L^{-1}(w))##. Given that we've decided on the dual basis, then

$$\tilde{T}_{ij} = \tilde{T}(e_i,e_j) = T(e_i,L(e_j))$$

$$T_i^{~~j} = T(e_i,e^j) = \tilde{T}(e_i,L^{-1}(e^j))$$

I'm assuming that I'll have to express ##L(e_j)## as a linear combination of dual basis vectors ##e^k##'s, and ##L^{-1}(e^j)## as a linear combination of basis vectors ##e_k##'s, but I'm at a loss on how to do that. That's primarily because in the example I gave above, I was able to express vector/covector components in terms of the other's components, but there's no indication on how to do that with vectors/covectors themselves. Any help would be appreciated.
If I understood you correctly, then you çan express in terms of components and then use multivibrator to find components. I

If I understood you correctly, then you çan express in terms of components and then use multivibrator to find components. I