Commutation relations of sl2

Question

Let $V$ be a vector space with basis $\{V_I\}$ indexed by all the subsets $I$ of $\{1,2,\cdots,n\}$ (including empty set). Thus $V$ has dimension $2^n$.

Question: V becomes a representation of Lie algebra $sl_2$ if we set $$e(v_I)= \sum_{{I \subset J}_{|J\setminus I|=1}}v_J,\quad f(v_I)= \sum_{{K \subset I}_{|I\setminus K|=1}}v_K , \quad h(v_I) = (2|I|-n)v_I$$

I know that I have to get $$[e,f]=h, \quad [h,e]=2e , [h,f]= -2f$$ and I started with $$[e,f](v_I) = e(f(v_I))- f(e(v_I))$$ to see if I will get the desired commutation relations but I end up getting $e,f]=0$ . Any help to bail me out will be much appreciated. I know if I see how one of the commutation relations work< I can figure out the remaining 2. Thanks.

Answer 1

The first and most obvious thing is to observe that both $e(f(v_I))$ and $f(e(v_I))$ are linear combinations of vectors $v_{I'}$ with $|I|=|I'|$. This is because $e$ adds one element to $I$, and $f$ removes one. Also, if $v_{I'}$ appears in either $e(f(v_I))$ or $f(e(v_I))$, then the symmetric difference $$ I\,\Delta\, I'=(I\setminus I')\cup (I'\setminus I) $$ is either the empty set (when $I=I'$), or it has exactly two elements, say $i\in I\setminus I'$ and $j\in I'\setminus I$.

• In the latter case we have no choice: $e$ must have added $j$ to $I$, $v_{I\cup\{j\}}$ appears in $e(v_I)$, and $f$ must have removed $i$ from $I\cup\{j\}$ meaning that $v_{I'}$ appears in $f(e(v_I))$ with coefficient $1$. But, the exact same thing must have happened in $e(f(v_I))$. Therefore $v_{I'}$ disappears from $[e,f](v_I)$.
• OTOH, when $I'=I$ we have more options. In $f(e(v_I))$ the action of $e$ can add any of the $n-|I|$ numbers not in $I$ to it. Then $f$ removes the added number. Consequently $v_I$ appears with coefficient $n-|I|$ in $f(e(v_I))$. Similarly, in $e(f(v_I))$ the operator $f$ can remove any of the $|I|$ elements from it as long as $e$ puts it back in. Therefore $v_I$ appears with coefficient $|I|$ in $e(f(v_I))$.

Putting the above bullets together we see that $$ [e,f](v_I)=(2|I|-n)v_I=h(v_I) $$ as claimed.

Answer 2

Could you please see what I am doing wrong below? Let say $V$ is a vector space with basis $$\{v_{\emptyset}, v_{\{1\}}, v_{\{2\}}, v_{\{1,2\}} \}$$ indexed by all the subsets $\{1,2\}$. This case $n=2$. I still had trouble seeing that $$[h,e](v_{\{1\}})=2e v_{\{1\}}.$$

I started by seeing what looks like. Observe that $$[h,e](v_{1})= h(e(v_{\{1\}}))- e(h(v_{\{1\}}))= h(v_{\{1\}}+ v_{\{1,2\}})-e((2(1) -2)v_{\{1\}})$$ $$= h(v_{\{1\}})+h( v_{\{1,2\}}) = 0 + [2(2)-2]v_{\{1,2\}} = 2 v_{\{1,2\}} \neq 2v_{\{1\}}$$

Thanks. I am sure if this part can be clarified, then I should be fine. My problem is I dont know how the summation given above works.