Questions about Lorentz Matrices and Lorentz Metric

Question

(I use the abstract index notation convention in this post)

In $\mathbb{R}^4$, denote the Lorentz Metric as $g_{\mu\nu}=$diag$(-1,1,1,1)$, then we can define the Lorentz Matrices to be all $4\times 4$ matrices $\Lambda^{\mu}_{\;\; \nu}$ such that $$ \Lambda^{a}_{\;\; \mu} g_{ab} \Lambda^b_{\;\; \nu}=g_{\mu\nu} $$ now, my First Question is, say we get a Symmetric $4\times 4$ matrix $X_{\mu\nu}$ (i.e. $X_{\mu\nu}=X_{\nu\mu}$), such that for any Lorentz Matrix $\Lambda$, one has $\Lambda^{a}_{\;\; \mu} X_{ab} \Lambda^b_{\;\; \nu}=X_{\mu\nu}$. Then can we conclude that there exists some $\alpha\in \mathbb{R}$ with $X_{\mu\nu}=\alpha \Lambda_{\mu\nu}$?

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If the above conjecture is ture, then my Second Question will be that say we get a Totally Symmetric tensor $T_{abcd}$ (i.e. $T_{abcd}$ is unchanged under any permutation of indices), such that for any Lorentz Matrix $\Lambda$, $$ \Lambda^{a}_{\;\;\mu}\Lambda^{b}_{\;\;\nu}\Lambda^{c}_{\;\;\sigma}\Lambda^{d}_{\;\;\rho} T_{abcd}=T_{\mu\nu\sigma\rho} $$ Then can we conclude that $T_{abcd}$ can be expressed as $$ T_{abcd}=\alpha(g_{ab}g_{cd}+g_{ac}g_{bd}+g_{ad}g_{bc}) $$ for some $\alpha\in \mathbb{R}$? (notice the RHS above is already completely symmetric).

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If all of the aboves are ture, then my Third Question is can we extend this result for $2n$-order tensor $T_{a_1 a_2\cdots a_{2n-1}a_{2n}}$?

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What I know now is that since the scalar matrix $-I$ is a Lorentz Matrix, then for any odd order tensor $T$ with the above property, $T$ will be zero.

And I guess these thoughts can help to illustrate the duality between the Lorentz Metric and the Lorentz Matrix.

Answer 1

Yes, the first conjectured fact is clearly true. Symmetry is not even necessary. $\eta:= diag(-1,1,1,1)$ henceforth.

PROPOSITION. Let us assume that $$\Lambda^t X \Lambda = X$$ for all $\Lambda \in SO(1,3)_+$ and some $X \in M(4,\mathbb{R})$. Then $$X = c\eta$$ for some $c\in \mathbb{R}$.

PROOF.

It is not difficult to prove that the (orthochronous proper) Lorentz group is closed with respect to the transposition operation, so that $$\Lambda \eta \Lambda^t = \eta$$ if and only if $$\Lambda^t \eta \Lambda = \eta\:.$$ Therefore, from $$\Lambda^t X \Lambda = X$$ we have $$\eta \Lambda^t X \Lambda = \eta X $$ $$\Lambda \eta \Lambda^t X \Lambda = \Lambda \eta X$$ $$\eta X \Lambda = \Lambda \eta X$$ We conclude that $\eta X$ commutes with the fundamental representation of $SO(1,3)_+$. Since it is irreducible, we have that $$\eta X = c I \tag{1}$$ for some $c \in \mathbb{R}$. Multiplying both sides of (1) with $\eta$ we have $$X = c\eta\:.$$ QED

Regarding your further conjectures, they are known facts (I do not remember the precise statement of your third conjecture), but the proofs are not so easy to find in the literature. In

Goodman R., Wallach N.R.: Symmetry, Representations, and Invariants, Graduate Texts in Mathematics, vol. 255. Springer, New York (2009)

and

Weyl, H.: The Classical Groups: Their Invariants and Representations. Princeton University Press, Princeton (1997)

you should find the proofs or ideas to prove those statements.

In this paper of mine and I. Khavkine we used those facts.

I referred my statement to $SO(1,3)_+$ instead of $O(1,3)$. One should pay attention to the use of $SO(1,3)_+$ vs $O(1,3)$. That is because $\epsilon_{abcd}$ is also invariant under the action of $SO(1,3)_+$. However it is not symmetric...