I Orthonormal basis expression for ordinary contraction of a tensor

Shirish
Messages
242
Reaction score
32
I'm reading Semi-Riemannian Geometry by Stephen Newman and came across this theorem:

Orthonormal Basis Expression for Ordinary Contraction: Let ##(V,g)## be a scalar product space with signature ##(\epsilon_1, \ldots, \epsilon_n)##, and let ##(e_1,\ldots,e_n)## be an orthonormal basis for ##V##. If ##A\in\mathcal{T}^1_s(V)## (where ##s\geq 2)## and ##l<s##, then the tensor ##C^1_l(A)\in\mathcal{T}^0_{s-1}(V)## is given by: $$C^1_l(A)(v_1,\ldots,v_{s-1})=\sum_i\epsilon_i\langle\mathcal{R}^{-1}_s(A)(v_1,\ldots,v_{l-1},e_i,v_{l+1},\ldots,v_{s-1}),e_i\rangle$$

For context, ##\mathcal{R}_s:Mult(V^s,V)\to\mathcal{T}^1_s## is the representation map, which acts like this:
$$\mathcal{R}_s(\Psi)(\eta,v_1,\ldots,v_s)=\eta(\Psi(v_1,\ldots,v_s))$$
I don't understand the importance of the relation that the author is trying to prove here - it seems kind of irrelevant. Judging from the name of the theorem, shouldn't it be sufficient to figure out ##C^1_l(A)(e_{j_1},\ldots,e_{j_{s-1}})##, i.e. the components of the contracted tensor w.r.t. different combinations of orthonormal basis vectors? And indeed this is what the author does in the first 2 lines of the proof.

But why are we going overboard to get the result in the theorem statement? Does it mean something significant?
 
Physics news on Phys.org
Shirish said:
But why are we going overboard to get the result in the theorem statement? Does it mean something significant?
It gives you the contraction evaluated at any vectors. Of course it is enough to know it for basis vectors, but this is more general and equally easy.
 
martinbn said:
It gives you the contraction evaluated at any vectors. Of course it is enough to know it for basis vectors, but this is more general and equally easy.
Sorry if I sound ignorant since I'm completely new to these topics. To me the most intuitive form of the general case is:
$$C^1_l(A)(v_1,\ldots,v_{s-1})=A(\theta^i,v_1,\ldots,v_{l-1},e_i,v_l,\ldots,v_{s-1})$$
where ##\theta^i## is the dual basis vector corresponding to ##e_i##.

I guess the RHS in the equation in the theorem is because we don't even want to see ##\theta^i##, so we use the elaborate construction to get rid of it and just have an RHS in terms of ##e_i##'s?
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top