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Consider the following matrices, $$H=\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right) \qquad X=\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right) \qquad Y=\left(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}\right) $$ Let $\rho:\mathfrak{su}(2)\rightarrow \text{End}(V)$ be a lie algebra homomorphism such that $\rho(X)^*=-\rho(X)$. Does this imply that $\rho(Y)^*=-\rho(Y)$?

I have been trying to use the commutation relations of $H,X,Y$ and the fact that $\rho$ is a lie algebra homomorphism to show this but I can't see to do it.

Am I overlooking something obvious here?

J. W. Tanner
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samlanader
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  • Those matrices are not in $\mathfrak{su}(2)$. I assume you mean $\rho: \mathfrak{su}(2){\mathbb C} \rightarrow End{\mathbb C} (V)$ (with $V$ a finite-dim. complex vector space)? – Torsten Schoeneberg Dec 07 '19 at 00:30
  • Also, could you write down a non-trivial (but easy) example of such a representation where $\rho(X)^\ast =-\rho(X)$? – Torsten Schoeneberg Dec 07 '19 at 00:42
  • Yes you are coorect I should have said $\rho: \mathfrak{su}(2){\mathbb C} \rightarrow End{\mathbb C}(V)$. If I understand correctly, to write donw a representation it is enough to write down what it does to the basis elements $H,X,Y$? I thought sending $H,X,Y$ to the pauli matrices might work but that doesn't respect the bracket. Now I am trying to map into $\mathbb{R}^3$ and use the cross product as the lie algebra bracket? Is this along the right line or could you hint at an easy example? – samlanader Dec 07 '19 at 14:46
  • @TorstenSchoeneberg the representation $\rho: \mathfrak{su}(2){\mathbb C}\rightarrow \mathfrak{su}(2){\mathbb C}$ given by $\rho(A)=iA$ seems to work. This certainly gives $\rho(Y)^*=-\rho(Y)$ – samlanader Dec 07 '19 at 16:06
  • But is this true for all representations ? – samlanader Dec 07 '19 at 19:11
  • That's not a representation, since $[iX, iY] \neq i[X,Y]$. In general, multiplication with a constant $c \neq 0,1$ is not a Lie algebra homomorphism unless the Lie algebra is abelian. – Torsten Schoeneberg Dec 08 '19 at 00:00
  • By the way, I assumed above and in my answer that $^\ast$ means hermitian transpose. But besides the proposed $\rho$ not being a representation, also $\pmatrix{0&i\0&0}^\ast =\pmatrix{0&0\-i&0} \neq \pmatrix{0&-i\0&0}$. Maybe you just meant complex conjugate with $^\ast$? The answer to that might be less trivial. – Torsten Schoeneberg Dec 08 '19 at 21:59

1 Answers1

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The assertion is true, but for a striking reason: If $\rho: \mathfrak{su}(2)_{\mathbb C} (\simeq \mathfrak{sl}_2(\mathbb C)) \rightarrow End_{\mathbb C}(V)$ is a (finite dimensional, complex) Lie algebra representation with $\rho(X)^\ast =-\rho(X)$, then $\rho$ is the trivial representation $\rho =0$.

Namely, it is well-known or easily proven that $\rho(X)$ must be nilpotent. But:

Lemma: Let $V$ be a finite-dimensional $\mathbb C$-vector space with hermitian inner product $\langle \cdot, \cdot\rangle$, and let $A \in End_{\mathbb C}(V)$ be nilpotent. Then $A^\ast =-A \Rightarrow A=0$.

Proof: Assume $A \neq 0$, then there is $0 \neq v_1 \in V$ with $v_0 := Av_1 \neq 0$ and $Av_0 =0$. So $$ 0 = \langle 0, v_1 \rangle =\langle Av_0, v_1 \rangle = \langle v_0, A^\ast v_1\rangle = \langle v_0, -Av_1 \rangle = -\langle v_0, v_0 \rangle \neq 0,$$ contradiction. (For different proofs and generalisations, compare e.g. $A$ is normal and nilpotent, show $A=0$.)

Now from $\rho(X)=0$ it follows easily (if one does not know it from the representation theory of $\mathfrak{sl}_2(\mathbb C)$) that $\rho(H)$, and a fortiori $\rho(Y)$, are $=0$.

Torsten Schoeneberg
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