Prove that tensor of the type $(1,1)$ which is invariant under orthogonal transformations of $\mathbb{R}^n$ is proportional to the tensor $\delta^{i}_{j}$.

Approach: Let $V=\mathbb{R}^n$ and $T:V\times V^*\to \mathbb{k}$ be a tensor of type $(1,1)$. Let $\{e_1,\dots,e_n\}$ and $\{\tilde{e}_1,\dots,\tilde{e}_n\}$ be two basis of $V$ such that matrix $C$ is transformation matrix from $(e)$ to $(\tilde{e})$. Then $\tilde{e}_j=c^l_je_l$ and $\tilde\varepsilon^i=d^i_k\varepsilon^k$ where by $\{\varepsilon^i\}$ and $\{\tilde{\varepsilon}^i\}$ I mean corresponding dual basis.

Let $\tilde{T}^i_j=T(\tilde{e}_j,\tilde{\varepsilon}^i)=T(c^l_je_l,d^i_k\varepsilon^k)=c^l_jd^i_kT(e_l,\varepsilon^k)=c^l_jd^i_kT^k_l$, where $c^i_j$ and $d^i_j$ are elements of $C$ and $C^{-1}$, respectively.

I was wondering what does mean that our tensor is invariant under orthogonal transformation?

I have the following hypothesis: maybe $T^i_j=c^l_jd^i_kT^k_l$ for any orthogonal matrix $C$?

Note that in the LHS I wrote $T^i_j$ instead of $\tilde{T}^i_j$.

Am I wrong? Anyway what is the definition of invariant tensor? Would be very grateful for any comments!

EDIT: Actually I solved this problem before I posted my question so let me write down my solution: since our tensor of type $(1,1)$ is invariant then $T^i_j=c^l_jd^i_kT^k_l$ where $c^i_j, d^i_j$ are elements of matrix $C$ and $C^{-1}$, respectively and $C$ is orthogonal matrix.

Let's take the following diagonal matrix $C_r$ s.t. $c_{rr}=-1$ and $c_{ii}=1$ for $i\neq r$. We see that $C_r$ is orthogonal and $C^{-1}=C$. Then one can show that $T^r_j=c^l_jd^r_kT^k_l=d^r_kT^k_jc^j_j=d^r_rT^r_jc^j_j$ and for $j\neq r$ we have $T^r_j=-T^r_j$ and it means that $T^r_j=0$. Since $r$ is arbitrary then it follows that $T^r_j=0$ for all $j\neq r$.

Consider the following orthogonal transformation given by its matrix $C$ such that $C$ is obtained from identity matrix by interchanging $i$th and $j$th columns. Then $$T^i_i=c^l_id^i_kT^k_l=d^i_kT^k_jc^j_i=d^i_kT^k_j=T^j_jd^i_j=T^j_j$$

So it shows that $T^i_j=c\delta^i_j$ where $c$ is constant.


1 Answer 1


Write $\widetilde{e}_j = A^i_{~j}e_i$ for some orthogonal matrix $(A^i_{~j})$. Then the dual relations are $\widetilde{e}^i = (A^\top)^i_{~j}e^j$. The relation between the components of $T$ relative to both bases is $$\widetilde{T}^i_{~j} = (A^\top)^i_{~k}A^\ell_{~j}T^k_{~\ell},$$but the assumption for the problem is that $T^i_{~j} = \widetilde{T}^i_{~j}$. Hit both sides with $A^p_{~i}$ to get $$ A^p_{~i}T^i_{~j} = A^p_{~i}(A^\top)^i_{~k}A^\ell_{~j}T^k_{~\ell} = \delta^p_{~k}A^\ell_{~j}T^k_{~\ell} = A^\ell_{~j}T^p_{~\ell}$$Because of this, the problem boils down, on the matrix level, to showing that if $T$ is a $n\times n$ matrix such that $AT = TA$ for all $A \in {\rm O}(n,\Bbb R)$, then $T$ is a multiple of ${\rm Id}_n$. This, in turn, is explained in this post.

  • $\begingroup$ Thanks a lot for your answer! But could you also take a look at my solution in my edit? $\endgroup$
    – RFZ
    Commented Apr 18, 2020 at 15:40
  • $\begingroup$ Could you explain why $T^i_j=\tilde{T}_j^i$? I have asked you about my edit but you did not answer $\endgroup$
    – RFZ
    Commented Apr 19, 2020 at 0:17
  • $\begingroup$ I don't have time to read all of it right now (I'll try later) but the second thing you asked is the invariance assumption. $\endgroup$
    – Ivo Terek
    Commented Apr 19, 2020 at 0:19
  • $\begingroup$ Could you explain in details what is invariance of tensor? $\endgroup$
    – RFZ
    Commented Apr 19, 2020 at 0:19

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