Suppose $\mathfrak g$ is a real finite dimensional Lie algebra. If I understand it correctly, Ado's Theorem states that there is a real vector space $V$ and an injective Lie algebra homomorphism $\pi:\mathfrak g \to \mathrm{End}(V)$.

My question is what the smallest dimension of the vector space $V$ is that is needed. I believe that if $\mathfrak g$ has only a trivial center, that then the answer is $\mathrm{dim}(\mathfrak g)$, because one can use the adjoint representation. But is this really the minimal one?


1 Answer 1


No, this need not be the minimal one. Let $\mathfrak{g}$ be an $n$-dimensional Lie algebra over a field $K$ of characteristic zero. Define an invariant of $\mathfrak{g}$ by $$\mu (\mathfrak{g}):=\min \{\dim_K (M) \mid M \text{ is a faithful $\mathfrak{g}$-module}\}.$$ For complex simple Lie algebras $\mathfrak{s}$, for example, we always have $\mu (\mathfrak{s})< \dim (\mathfrak{s})$, except for type $E_8$. Here is a small table over $\Bbb C$.

\begin{array}{c|c|c} \mathfrak{g} & \dim (\mathfrak{g}) & \mu(\mathfrak{g}) \\ \hline A_n, \; n\ge 1 & (n+1)^2-1 & n+1 \\ B_2 & 10 & 4 \\ B_n, \; n\ge 3 & 2n^2+n & 2n+1 \\ C_n, \; n\ge 3 & 2n^2+n & 2n \\ D_n, \; n\ge 4 & 2n^2-n & 2n \\ E_6 & 78 & 27 \\ E_7 & 133 & 56 \\ E_8 & 248 & 248 \\ F_4 & 52 & 26 \\ G_2 & 14 & 7 \\ \end{array}

In general, the question about the minimal dimension is quite difficult and we only have lower and upper bounds. For references see the papers (my papers and other papers) in this article.

For example, let $\mathfrak{g}$ be a nilpotent Lie algebra of dimension $n$ and nilpotency class $k$. Denote by $p(j)$ the number of partitions of $j$ and let $$p(n,k)=\sum_{j=0}^{k}\binom{n-j}{k-j}p(j).$$ Then $\mu(\mathfrak{g})\le p(n,k)$. If $k=n-1$, with $n\ge 3$, then $n\le \mu(\mathfrak{g})$.

  • $\begingroup$ Thanks for the quick answer! I am a little bit confused by the definition of $\mu(\mathfrak g)$, are (faithful) $\mathfrak g$-modules the same thing as (faithful) linear representations? If yes, how are the dimensions related? Sorry, I am not so familiar with abstract algebra. $\endgroup$
    – Lau
    Commented Jun 23, 2022 at 14:39
  • 1
    $\begingroup$ Yes, faithful module means injective linear representation. So exactly what you have asked. $\endgroup$ Commented Jun 23, 2022 at 14:40
  • $\begingroup$ Totally off-topic here, but another reminder that $B_2$ "should" be called $C_2$ , as proposed in mathoverflow.net/q/52323/27465. $\endgroup$ Commented Jun 23, 2022 at 16:47
  • $\begingroup$ Thanks for the clarification. I'm mainly interested in real Lie algebras (in particular, subalgebras of $\mathfrak{su}(N)$). I surely can complexify my (real) Lie algebra and then apply the result in your paper (if that algebra is reductive) but that's in general not optimal right? $\endgroup$
    – Lau
    Commented Jun 23, 2022 at 17:56
  • $\begingroup$ Most of the results in the literature are also valid for real Lie algebras, or can be easily adapted, so there is no problem. $\endgroup$ Commented Jun 23, 2022 at 18:50

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