MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the lowest dimension of a faithful ordinary representation (as compared with projective representation) of the projective unitary group $\rm{PU}(d)$? Is it $d^2-1$?

share|cite|improve this question
It helps to explain the notation and setting more explicitly. Aside from that, it's unclear how you distinguish a projective representation from an ordinary one when the group itself is constructed from the usual unitary group by factoring out its center. – Jim Humphreys Feb 26 '13 at 19:21

If $G\subset GL_n({\mathbb C})$ is a compact Lie group, and $G({\mathbb C})$ is the Zariski closure of $G$ in $GL_n$, then every continuous representation of $G$ extends uniquely to an algebraic representation of $G({\mathbb C})$. Consequently, if we take $G=PU(d)$, then $G({\mathbb C})=PGL(d)=SL(d) \quad modulo \quad centre$, and the question of minimality of irreps for $G$ becomes one for $PGL(d)$. By the Borel -Weil Theorem, every irreducible representation of $SL(d)$ is of the form $V_{\chi}=Ind_B ^G (\chi)$ where $B$ is the Borel subgroup of upper triangular matrices, and $\chi$ is an anti-dominant character. In order that this representation descend to $PGL(d)$ it is necessary and sufficient that $\chi$ be trivial on the centre of $SL(d)$.

By Using the Weyl dimension formula, one can then see that the smallest $V_{\chi}$ is the adjoint representation. To see this, let $\lambda$ be the highest weight of the representation $V_{\chi}$. In the usual notation,

$$\lambda =(m_1,\cdots, m_{d-1})$$ where $m_i$ are non-negative integers which are decreasing. Set $a_i=m_i+d-i$. The dimension of this representation $V_{\chi}$ is the product $$\prod _{1\leq i< j\leq d} \frac{a_i-a_j}{i-j}.$$ Since we have that $\lambda $ is trivial on the centre of $SU(d)$, it follows that $m_1+\cdots+m_{d-1}$ is divisible by $d$. Hence $m_1\geq 2$. This is easily seen to imply that the dimension of $V_{\chi}$ is at least $d^2-1$ with equality if and only if $m_1=2$ and $m_2=\cdots =m_{n-1}=1$. That is, $\lambda$ is the highest weight of the adjoint representation.

share|cite|improve this answer
Making the "Zariski density arguments" more explicit would be helpful. since the question seems to involve real Lie groups and not just complex ones. Also, you don't actually have to invoke the Borel-Weil Theorem for this limited purpose, while the Weyl dimension formula is intended at first for compact (or complex) semisimple groups. The question itself is stated only in a sketchy way. – Jim Humphreys Feb 26 '13 at 19:29
@Humphreys:Thank you. I have put in a few words on Zariski density. – Venkataramana Feb 27 '13 at 0:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.