# About invariant of ortogonal group

For an arbitrary $k$-linear form $P$ on $R^n$ of the form: $$P(x_1,..,x_k)=\sum_{i_1,...,i_k=1}^n a_{i_1,...,i_k}x_{1i_1} ...x_{ki_k}$$ we put $$I(P):=\sum_{i_1,...,i_k=1}^n a_{i_1,...,i_k}^2.$$ Is this quantity invariant with respect to ortogonal group? In other words, let $P$ be $k$ linear form, $f\in O(n)$ and $P_f(x_1,...,x_k):=P(f(x_1),...,f(x_k))$. Is it then $$I(P)=I(P_f) ?$$

I know that it is true for $k=2$. In that case $P(x_1,x_2)=x_1^TAx_2$, $P_f(x_1',x_2')= x_1'^TC^TACx_2'$ (in matrix denotation). Then $Tr(A A^T)=\sum_{i_1,i_2}^n a_{i_1i_2}^2$ and $Tr((C^TAC)(C^TAC)^T)=Tr(C^{-1}AA^TC)=Tr(AA^T)=\sum_{i_1,i_2}^n a_{i_1i_2}^2$ ( since $C^{-1}=C^T$).

Thanks.

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Yes. Put together, the points below entail $I(P_f)=I(P)$:
• Observe that if $J_\sigma(x_1,\cdots,x_k)=P(x_{\sigma(1)},\cdots,x_{\sigma(k)})$, we have $I(J_\sigma)=I(P)$.
• Show that if $Q(x_1,\cdots,x_k)=P(f(x_1),x_2,x_3,\cdots,x_k)$, we have $I(Q)=I(P)$.