Prove that there exist a tensor whose components relative to the product basis of $\pmb{any}$ basis are the values of the generalized kronecker delta

Question

ACHTUNG!!!

To follow I adopt the Einstein notation!!!

Definition

The dual space $V^*$ of a vector space $V$ over the field $\Bbb F$ is the vector space of the linear functions from $V$ to $\Bbb F$.

Definition

Give a basis $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ of a vector space $V$ the dual set of $\mathcal E$ is the set $\mathcal E^*:=\{\vec e^{\,1},...,\vec e^{\,n}\}$ of $V^*$ whose elements are defined through the equality $$ \vec e^{\,i}(\vec e_j)=\delta^i_j $$ for each $i,j=1,..,n$.

Theorem

If $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ is a basis for a vector space $V$ then the dual set $\mathcal E^*$ of $\mathcal E$ as above defined is a basis of the dual space $V^*$ of $V$.

Theorem

Let be $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ and $\mathcal B:=\{b_1,..,b_n\}$ two basis of a vector space $V$ so that $$ \vec b_k=B^s_k\vec e_s\,\,\,\text{and}\,\,\,\vec e_q=E^k_q\vec b_k $$ for each $k,q=1,...,n$. So if this holds then $$ \vec b^{\,\,q}=E^q_k\vec e^{\,\,k}\,\,\,\text{and}\,\,\,\vec e^{\,\,q}=B^q_k\vec e^{\,\,k} $$ for each $k,q=1,...,n$ and moreover $B^q_kE^k_s=\delta^q_s$ for each $q,s=1,...,n$

Definition

If $V_1,...,V_s$ is a collection of vector spaces over the same field $\Bbb F$ then a $s$-linear function is a function $f$ form $V_1\times...\times V_s$ in $\Bbb F$ that is linear in each of its variables while the other variables are held costant. In particular if $V_i=V^*$ for some $r<s$ and $V_i=V$ otherwise we say that the function $f$ is a tensor of contravariant order $p:=r$ and covariant order $q:=(s-r)$.

Theorem

If $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ is a basis of $V$ then a tensor $\tau$ of contravariant order $p$ and covariant order $q$ is completely defined by the equality $$ T^{i_1,...,i_p}_{j_1,..,j_q}:=\tau\big(\vec e^{\,i_1},...,\vec e^{\,i_p},\vec e_{j_1},...\vec e_{j_q}\big) $$ for $i_1,...,i_p,j_1,...,j_q=1,...,n$. Furthermore if $\mathcal B:=\{\vec b_1,..,\vec b_n\}$ is another basis of $V$ then $$ \tau(\vec b^{\,i_1},..,\vec b^{\,i_p},\vec b_{j_1},...,\vec b_{j_q})=E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_q}_{j_q}T^{h_1,...,h_p}_{k_1,....,k_q} $$ for each $i_1,...,i_p,j_1,...,j_q=1,...,n$.

Now let be $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ a basis of a vector space $V$ so that we can define a tensor $\tau$ of contravariant and covariant order p letting that $$ \tau(\vec e^{\,i_1},...,\vec e^{\,i_p},\vec e_{j_1},..,\vec e_{j_p})=\delta^{i_1,...,i_p}_{j_1,..,j_p} $$ for each values of the indices $i_1,...,i_p,j_1,...,j_p$ from $1$ to $n$. So I ask to prove that for any vector space and for any natural number $p$ there exist a tensor $\tau$ of order $p$ whose components relative to the product basis of any basis are the values of the generalized kronecker delta as here defined, that is more explicitly I ask to prove that if $\tau$ is the tensor above defined and if $\mathcal B:=\{\vec b_1,..,\vec b_n\}$ is another basis for the space $V$ above defined then $$ \tau(\vec b^{\,i_1},...,\vec b^{\,i_p},\vec b_{j_1},..,\vec b_{j_p})=E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}\delta^{h_1,...,h_p}_{k_1,...,k_p} $$ for each values of the indices $h_1,...,h_p,k_1,...,k_p$.

So could someone help me, please?

Answer 1

First of all we remember that if $\epsilon^{i_1,...,i_k}$ is the Levi-Civita symbol and if $A$ is a square matrix of order $n$ then $$ \det A=\epsilon^{i_1,...,i_n}A^{i_1}_1....A^{i_n}_n=\epsilon_{i_1,...,i_n}A^1_{i_1}....A^n_{i_n}\,\,\,\text{and}\,\,\,\epsilon^{i_1,...,i_k}\epsilon_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k} $$ and moreover as showed here it holds the equality $$ \delta^{i_1,...,i_k}_{j_1,...,j_k}=\text{det}\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_k}\\ \vdots&\ddots&\vdots\\ \delta^{i_k}_{j_k}&\cdots&\delta^{i_k}_{j_k}\end{pmatrix} $$ for each $k\in\Bbb N$ and for each value of the indices $i_1,...,i_k$ and $j_1,...,j_k$. So remembering this we observe that $$ E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}\delta^{h_1,...,h_p}_{k_1,...,k_p}=\epsilon^{h_1,...,h_p}E^{i_1}_{h_1}...E^{i_p}_{h_p}\epsilon_{k_1,...,k_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}=\\ \det\begin{pmatrix}E^{i_1}_1&\cdots& E^{i_1}_p\\ \vdots&\ddots&\vdots\\ E^{i_p}_1&\cdots&E^{i_p}_p\end{pmatrix}\det\begin{pmatrix}B^{1}_{j_1}&\cdots& B^1_{j_p}\\ \vdots&\ddots&\vdots\\ B^{p}_{j_1}&\cdots& B^p_{j_p}\end{pmatrix}=\\ \det\begin{pmatrix}\begin{pmatrix}E^{i_1}_1&\cdots& E^{i_1}_p\\ \vdots&\ddots&\vdots\\ E^{i_p}_1&\cdots&E^{i_p}_p\end{pmatrix}&\begin{pmatrix}B^{1}_{j_1}&\cdots& B^1_{j_p}\\ \vdots&\ddots&\vdots\\ B^{p}_{j_1}&\cdots& B^p_{j_p}\end{pmatrix}\end{pmatrix}=\\ \det\begin{pmatrix}E^{i_1}_{s^1_1}B^{s^1_1}_{j_1}&\cdots&E^{i_1}_{s^1_p}B^{s^1_p}_{j_p}\\ \vdots&\ddots&\vdots\\ E^{i_p}_{s^p_1}B^{s^p_1}_{j_1}&\cdots&E^{i_p}_{s^p_p}B^{s^p_p}_{j_p}\end{pmatrix}=\det\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_p}\\ \vdots&\ddots&\vdots\\ \delta^{i_p}_{j_1}&\cdots&\delta^{i_p}_{j_p}\end{pmatrix}=\delta^{i_1,...,i_p}_{j_1,...,j_p} $$ for each values of the indices $i_1,...,i_p$ and $j_1,..,j_p$.

Answer 2

Warning: The following uses the invariant definition of a tensor product which can be found in any good book of algebra but is often written in such a way that it hides the essential clarity of the definition.

Let $V \in FinVect$.

Define $\delta:=e^i \otimes \e_i$. Here a sum is implied by Einsteims convention - said to be, by some (ironically), his greatest contribution to maths.

Let $v \in V$

Then $\delta.v = (e^i.v)e_i = v^i.e_i = v$ Hence we see that $\delta = Id: V \rightarrow V$

This is the invariant definition of the Kronecker delta in its simplest form and is simply the identity, which is obviously invariant. To check that it is, let us take components, then

$\delta^i_j$ $ = e^i(\delta)e_j$ $= e^i(e^k\otimes e_k)e_j$ $= e^i((e^k.e_j).e_k)$ $= e^i(\delta^k_j.e_k)$ $= \delta^k_j(e^i.e_k)$ $=\delta^k_j.\delta^i_k$ $=\delta^i_j$

And which matches.