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Background: Let $$V:=\{A\in M_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$ be the vector space of $3\times 3$ real matrices with vanishing trace, and let $[\cdot,\cdot]:V\times V\to V$ be defined by $$[A,B]=AB-BA$$ Note that this is well defined since $\text{trace}(AB-BA)=\text{trace}(AB)-\text{trace}(AB)=0$ for any matrices $A,B$. Finally, let $$\hat{V}:=\text{span}\{[A,B]:A,B\in V\}$$

Question: I have to give an isomorphism $$\varphi:(V,[\cdot,\cdot])\to(\hat{V},[\cdot,\cdot]),$$ i.e. a bijective map such that $\varphi([A,B])=[\varphi(A),\varphi(B)]$, $\forall A,B\in V$.

I am aware that this is related to Lie algebra; namely $V=\mathfrak{sl}(3,\mathbb{R})$. However, I did not study Lie algebra yet. I am suppose to be able to prove it without referring to any well-known theorem of Lie algebra (unless I prove it first).

I used a very computational argument. I am wondering if

  1. My argument is correct.
  2. There is a more fundamental way to prove it, i.e. without having to tediously compute all possible $[X_i,X_j]$ from a given basis.

Atempt: Let

$$ X_1 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right),\quad X_2 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right),\quad X_3 = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right),$$ $$\quad X_4 = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right),\quad X_5 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right),\quad X_6 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right),$$ $$\quad X_7 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right),\quad X_8 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) $$ Then, $\{X_1,\ldots,X_8\}$ forms a basis for $V$, and we find $$[X_1,X_2]=0,\quad[X_1,X_3]=2X_3,\quad[X_1,X_4]=X_4,\quad[X_1,X_5]=-X_5$$ $$[X_1,X_6]=-2X_6,\quad[X_1,X_7]=-X_7,\quad[X_1,X_8]=X_8,\quad[X_2,X_3]=X_3$$ $$[X_2,X_4]=2X_4,\quad[X_2,X_5]=X_5,\quad[X_2,X_6]=-X_6,\quad[X_2,X_7]=-2X_7$$ $$[X_2,X_8]=-X_8,\quad[X_3,X_4]=0,\quad[X_3,X_5]=X_4,\quad[X_3,X_6]=X_1$$ $$[X_3,X_7]=-X_8,\quad[X_3,X_8]=0,\quad[X_4,X_5]=0,\quad[X_4,X_6]=-X_5$$ $$[X_4,X_7]=X_2,\quad[X_4,X_8]=X_3,\quad[X_5,X_6]=0,\quad[X_5,X_7]=X_6$$ $$[X_5,X_8]=X_2-X_1,\quad[X_6,X_7]=0,\quad[X_6,X_8]=-X_7,\quad[X_7,X_8]=0$$ Thus, we have $$ \begin{array}{cccc} X_1=[X_3,X_6],& X_2=[X_4,X_7],& X_3=[X_2,X_3],& X_4=[X_1,X_4] \\ X_5=[X_2,X_5],& X_6=[X_5,X_7],& X_7=[X_7,X_1],& X_8=[X_7,X_3] \end{array} $$ which shows that $V$ is a subspace of $\hat{V}$, and since $\hat{V}$ is clearly a subspace of $V$, then $V$ and $\hat{V}$ are in fact the same vector space. Thus, we can take $\varphi$ to be simply the inclusion map. Indeed, $\varphi$ is then clearly bijective and we have $$\varphi([A,B])=[A,B]=[\varphi(A),\varphi(B)]$$

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I think your argument is correct, but you can simplify it a bit. Let $E_{ij}$ denotes the matrix whose $(i,j)$-th entry is $1$ and all other entries are zero. Then for any three distinct indices $i,j,k$, we have \begin{align*} E_{ii}-E_{jj} &= [E_{ij},E_{ji}],\\ E_{ij} &= [E_{ik},E_{kj}]. \end{align*} This shows that the subspace traceless matrices $V$ lies inside the span $\widehat{V}$ of all commutators of pairs of traceless matrices. Note that this proof works for every $n\ge3$.

Remark. Actually, the set $\mathcal{C}_0:=\{AB-BA: \operatorname{trace}(A)=\operatorname{trace}(B)=0,\ A,B\in M_n(\mathbb{R})\}$ is identical to $\mathcal{C}:=\{AB-BA: A,B\in M_n(\mathbb{R})\}$. (That $\mathcal{C}_0\subseteq\mathcal{C}$ should be clear; $\mathcal{C}\subseteq\mathcal{C}_0$ because $AB-BA=\widehat{A}\widehat{B}-\widehat{B}\widehat{A}$, where $\widehat{A}:=A-\frac1n\operatorname{trace}(A)I$ and $\widehat{B}:=B-\frac1n\operatorname{trace}(B)I$ are traceless.) Since the set $\mathcal{C}$ of all commutators is known to identical to the subspace of traceless matrices (I have given an elementary proof here), the "$\operatorname{span}$" in your problem statement can actually be removed. In short, $V=\mathcal{C}=\mathcal{C}_0=\operatorname{span}(\mathcal{C}_0)=\widehat{V}$. But certainly, with the "$\operatorname{span}$", the problem is much easier to solve.

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I didn't check all your computations, but the idea is certainly right. You can reduce the amount of computation by noticing that $X_2$ is just $X_1$ with the coordinates relabeled (the second row and column are switched with the third), and that all of $X_4$ through $X_8$ are likewise obtainable from $X_3$ by suitable relabeling. So, once you've obtained $X_1$ and $X_3$ as commutators $[A,B]$, you don't have to worry about the rest, as they could be obtained by applying the same relabeling to $A$ and $B$. (To sound fancy, one can say "conjugation by a permutation matrix" instead of "relabeling", but I find it easier to think in terms of relabeling the coordinates.)

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