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 $d^21$?

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 antidominant 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, $$\lambda =(m_1,\cdots, m_{d1})$$ where $m_i$ are nonnegative integers which are decreasing. Set $a_i=m_i+di$. The dimension of this representation $V_{\chi}$ is the product $$\prod _{1\leq i< j\leq d} \frac{a_ia_j}{ij}.$$ Since we have that $\lambda $ is trivial on the centre of $SU(d)$, it follows that $m_1+\cdots+m_{d1}$ is divisible by $d$. Hence $m_1\geq 2$. This is easily seen to imply that the dimension of $V_{\chi}$ is at least $d^21$ with equality if and only if $m_1=2$ and $m_2=\cdots =m_{n1}=1$. That is, $\lambda$ is the highest weight of the adjoint representation. 

