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I want to find $\mathrm{Z}(\mathfrak{gl}(2,\Bbb F))$ where the Lie bracket is $[X,Y]=XY-YX$

So then this will depend on the field, but no harm in direct computation for arbitrary matrices: $$x=\begin{bmatrix}a&b\\c&d\end{bmatrix},y=\begin{bmatrix}\alpha&\beta\\\gamma&\delta\end{bmatrix}$$ $$[x,y]=\begin{bmatrix}a\alpha+b\gamma-\alpha a - \beta c&a\beta+b\delta-\alpha b-\beta d\\c\alpha+d\gamma -\gamma a - \delta c&c\beta + d\delta - \gamma b-\delta d\end{bmatrix}$$

I want to find $x\in\mathfrak{gl}(2,\Bbb F)$, $[x,y]=0,\forall y$

In $\Bbb C$ or $\Bbb R$, the only possible elements are $\begin{bmatrix}0&0\\0&0\end{bmatrix},I$

In $\Bbb Z_2$, the top left position gives us $b=c=0$, so $$[x,y]=\begin{bmatrix}a\alpha-\alpha a&a\beta-\beta d\\d\gamma -\gamma a & d\delta -\delta d\end{bmatrix}$$

That's easier to handle and we get $d-a=a-d=0$, which means the centre is:

$$Z(\mathfrak{gl}(2,\Bbb Z_2))=\left\{\begin{bmatrix}0&0\\0&0\end{bmatrix},I\right\}$$

How would I go about checking the centre for all fields? Will this always be the same?

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  • $\begingroup$ I just made the question about gl(2,F) since I accidentally forgot to make them traceless, and I am interested regardless. $\endgroup$ – So many hats Aug 14 '15 at 12:14
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A consideration of $\mathfrak{gl}(n,\Bbb F)$:

Let $E_{ij} = e_i e_j^T$ denote the matrix with a $1$ in the $i,j$ entry. Let $A$ be a matrix with entries $a_{ij}$. We have $$ AE_{ij} = (Ae_i)e_j^T\\ E_{ij}A = e_i(e_j^TA) $$ Now, if $A$ is in the center, we must have for every $p,q$: $$ e_{p}^T[A,E_{ij}]e_q = 0 \implies\\ e_p^T\left((Ae_i)e_j^T - e_i(e_j^TA) \right)e_q = 0 \implies\\ (e_p^T A e_i)(e_j^Te_q) - (e_p^Te_i)(e_j^TAe_q) = 0 \implies\\ a_{pi}\delta_{jp} - \delta_{pi} a_{jq} $$ where $\delta$ denotes the Konecker delta.

By choosing different $i,j,p,q \in \{1,\dots,n\}$, you can deduce that $a_{ij} = 0$ when $i \neq j$, and $a_{ii} = a_{jj}$ for each $i,j$.

In other words, the only elements of the center are the multiples of $I$. Note that this computation involves no division by a coefficient, and so it applies to all fields.

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  • $\begingroup$ This is actually really amazing, thank you!!! $\endgroup$ – So many hats Aug 14 '15 at 13:33
  • $\begingroup$ The same argument can be applied to $GL(n,\Bbb F)$ by considering $I + E_{ij}$. This argument new argument fails, however, in fields of characteristic $2$. $\endgroup$ – Omnomnomnom Aug 14 '15 at 13:40
  • $\begingroup$ And you're welcome $\endgroup$ – Omnomnomnom Aug 14 '15 at 13:40
  • $\begingroup$ @Omnomnomnom I know that this is an old question/answer, but I'm trying to understand it. Why did we choose to work with $E_{ij}$ in the first place? Some other posts talk about it "probing" the matrix $A$, but I don't understand what that means. $\endgroup$ – Irregular User Jan 26 '17 at 2:14
  • $\begingroup$ @IrregularUser here's a very precise answer: $[\cdot,\cdot]$ is a bilinear function over the vector space $gl(2,\Bbb F)$. The matrices $E_{ij}$ form a convenient basis of $gl$, since the elements of this basis have mostly $0$s as entries. $\endgroup$ – Omnomnomnom Jan 26 '17 at 2:20
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Since the Lie algebra $\mathfrak{g}=\mathfrak{gl}(2,K)$ is reductive, it is the direct sum of the commutator ideal and the center, i.e., we have $$ \mathfrak{gl}(2,K)=\mathfrak{sl}(2,K)\oplus Z(\mathfrak{gl}(2,K)). $$ Since $\dim \mathfrak{sl}(2,K)=3$, we obtain $\dim Z(\mathfrak{gl}(2,K))=1$. Because $I$ is in the center, it is a generator of the $1$-dimensional vector space, i.e., $Z(\mathfrak{gl}(2,K))=\{\alpha I \mid \alpha \in K \}$.

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