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Assume that $char(\mathbb{k}) = p > 3$ and let $W(1)$ be the Witt algebra over $\mathbb{k}$.

Recall that $W(1) = Der(A)$ where $A = k[t]/(t^p)$, a truncated polynomial ring over $\mathbb{k}$.

We know that as a vector space over $\mathbb{k}$ the Lie algebra $W(1)$ has basis

$\{e_i | − 1 ≤ i ≤ p − 2\}$

where $e_i = t^{i+1} \frac{d}{dt}$, and the Lie bracket in $W(1)$ has the property that

$[e_i , e_j ] = (j − i)e_{i+j}$ if $i + j ≤ p − 2$, and $[e_i , e_j ] = 0$ if $i + j ≥ p − 1$.

QUESTION:

Use the above formula for $[e_i , e_j ]$ with $i = 0$ to determine the eigenvalues and the eigenspaces of the linear operator $(ad\space e_0) ∈ gl(W(1))$, where $e_0 = t\frac{d}{dt}$.

I am really struggling to see how to approach this problem. Is there an obvious matrix representation of $(ad\space e_0)$ (where $(ad\space e_0)(e_j)=[e_0,e_j]$) that I am missing that would enable calculation of eigenvalues, or am I looking at this problem the wrong way?

user3450230
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1 Answers1

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You are looking for eigenvalues and eigenspaces of the linear operator $T=ad\;e_0$ (where $T:W(1) \to W(1)$). So you're looking for non-zero vectors, $v \in W(1)$, and scalars, $\lambda$, such that $T(v)=\lambda v$.

You have that $T(e_j) = (ad\;e_0)(e_j) = [e_0,e_j] = (j-0)e_{0+j} = je_j$.

This means that each $e_j$ is an eigenvector (with corresponding eigenvalue $j$).

Since the $e_j$'s form a basis for $W(1)$, this means you have a basis of eigenvectors ($ad\;e_0$ is diagonalized by this basis) and its eigenspaces are just the one dimensional subspaces of $W(1)$ spanned by the $e_j$'s ($j=-1\dots p-2$).

Bill Cook
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