# Find $\alpha \in \mathbb{R}$ such that there exists a Lie algebra representation of $\mathfrak{sl}_2 (\mathbb{R})$ on $C^\infty(\mathbb{R})$

I am trying to get started with the following exercise in relation to representation theory:

Define differential operators $$X, Y, Z_\alpha, \alpha \in \mathbb{R}$$, on $$C^\infty(\mathbb{R})$$ by: $$X f(t) = f''(t)$$ $$Y f(t) = t^2 f(t)$$ $$Z_\alpha f(t) = t f'(t) + \alpha f(t)$$ For $$f \in C^\infty(\mathbb{R})$$. Find $$\alpha \in \mathbb{R}$$ such that there exists a Lie algebra representation of $$\mathfrak{sl}_2 (\mathbb{R})$$ on $$C^\infty(\mathbb{R})$$ with image equal to span$$_\mathbb{C} \lbrace X , Y , Z_\alpha \rbrace$$ (HINT: Find an appropriate basis of $$\mathfrak{sl}_2 (\mathbb{R})$$ which is mapped to suitable complex multiples of $$X, Y , Z_\alpha$$ under the representation)

Even though there is a hint I am having some difficulties getting started with this problem. How do I do this?

We have the operators $$Xf = f'', \quad \quad Yf = t^2 f, \quad \quad Z_\alpha f = t f' + \alpha f$$

Now we compute the three (up to sign) possible commutators. First $$X$$ and $$Y$$:

\begin{aligned} \, [X, Y]f &= XYf - YXf \\ &= (t^2f)'' - t^2f'' \\ &= t^2 f'' + 4tf' + 2f - t^2f'' \\ &= 4tf' + 2f \\ &= 2 Z_{2} f \end{aligned}

Now $$Z_\alpha$$ and $$X$$:

\begin{aligned} \, [Z_\alpha, X]f &= Z_\alpha X f - X Z_\alpha f \\ &= tf''' + \alpha f'' - (tf' + \alpha f)'' \\ &= tf''' + \alpha f'' - (2f'' + tf''' + \alpha f'') \\ &= -2f'' \\ &= -2Xf \end{aligned}

and finally $$Z_\alpha$$ and $$Y$$:

\begin{aligned} \,[Z_\alpha, Y] &= Z_\alpha Y f - Y Z_\alpha f \\ &= t(t^2 f)' + \alpha(t^2 f) - t^2(tf' + \alpha f) \\ &= 2t^2 f + t^3 f' + \alpha t^2 f - t^3 f' - \alpha t^2 f \\ &= 2t^2 f \\ &= 2Yf \end{aligned}

Therefore, setting $$Z = -2Z_2$$ we have the commutator relations \begin{aligned} \,[Y, X] &= Z \\ [Z, X] &= 2X \\ [Z, Y] &= -2Y \end{aligned}

which are precisely the commutator relations of $$\mathfrak{sl}_2$$. If we define $$\mathfrak{sl}_2$$ to be the three-dimensional Lie algebra with basis $$e, f, h$$ and Lie bracket \begin{aligned} \,[e, f] &= h \\ [h, e] &= 2e \\ [h, f] &= -2f \end{aligned}

then the linear map $$\varphi$$ defined by $$\varphi(e) = X$$, $$\varphi(f) = Y$$, and $$\varphi(h) = Z$$ is a Lie algebra homomorphism, and hence the operators $$X, Y, Z$$ define a representation of $$\mathfrak{sl}_2$$.

• So $[X, Y] f = 2 Z_2$ thus it would perhaps be nice if $\alpha = 2$ is working. However, I am still not sure how this will help me to go forward?
– Miep
Apr 13, 2021 at 9:35
• @Miep Try the other two commutators as well. You're basically just looking to change basis amongst the $\{X, Y, Z_{\alpha}\}$ such that you end up with three operators obeying the $\mathfrak{sl}_2$ relations: $[e, f] = h$, $[h, e] = 2e$ and $[h, f] = -2f$. Apr 13, 2021 at 10:35
• I have computed the other two as well and I can see how it will help me in choosing a representation $\rho$ and showing that it is in fact a representation for the Lie algebra. However, I am not so sure when it comes to choosing a basis and changing it such that it is suitable.
– Miep
Apr 13, 2021 at 17:21
• @Miep It seems that taking $-2Z_2$ as the basis of the Cartan gives the usual presentation of $\mathfrak{sl}_2$. Apr 14, 2021 at 14:59