For a complex simple Lie algebra $\frak{g}$, which of its finite dimensional irreducible representations give non-faithful representations of the corresponding simply-connected compact Lie group.


More specifically, can somebody point me to a table detailing, for each series, those dominant weights for faithfulness fails.


**Edit**: To make my question clearer, I am asking in relation to the answer of this [question](https://math.stackexchange.com/questions/310296/when-can-you-build-up-all-representations-from-the-fundamental-and-antifundament), which asks when can you build up all representations from the fundamental and antifundamental ones? It is answered that, for an irreducible Lie algebra representation $V$, when the corresponding representation of $G$ is faithful, any other irreducible Lie algebra can be found in a tensor product $V^{\otimes k}$, for sufficiently high $k$. The answer is qualified with the following comment: 



> 
> One mild warning: there is an obvious representation of $\frak{}()$ which is not a faithful representation of the corresponding simply-connected compact Lie group when $n\geq 3$
> 
> 
> 


Is the "obvious representation" of $\frak{so}(n)$ the only representation for which this happens, or are there others?



Non-faithful irreducible representations of simple Lie groups
Let $G\_{sc}$ be as in the answer by Victor Protsak and let $\varpi\_1$, $\ldots$, $\varpi\_l$ be the fundamental dominant weights. 


Let $\lambda$ be a dominant weight and write $\lambda = \sum\_{i = 1}^l a\_i \varpi\_i$ for $a\_i \in \mathbb{Z}\_{\geq 0}$.


Then the irreducible representation of $G\_{sc}$ with highest weight $\lambda$ is faithful precisely in the following cases:


* Type $A\_{l}$ ($l \geq 1$): $\gcd(l+1, a\_1+2a\_2+\cdots+la\_l) = 1$.
* Type $B\_l$ ($l \geq 2$): $a\_l$ is odd.
* Type $C\_l$ ($l \geq 2$): $a\_1 + a\_3 + a\_5 + \cdots$ is odd.
* Type $D\_l$ ($l \geq 4$): $l$ is odd and $a\_{l-1} + a\_l$ is odd
* Type $G\_2$: always
* Type $F\_4$: always
* Type $E\_6$: $a\_1 - a\_3 + a\_5 - a\_6$ is not divisible by $3$.
* Type $E\_7$: $a\_2 + a\_5 + a\_7$ is odd
* Type $E\_8$: always


This can be determined by a direct computation. The kernel of any irreducible representation of $G\_{sc}$ lies in $Z(G\_{sc})$, which is finite. Furthermore, you can describe $Z(G\_{sc})$ explicitly and compute the action of any $z \in Z(G\_{sc})$ in an irreducible representation (it is of course always multiplication by some scalar). See for example Chapter 3, Lemma 28 in "Lectures on Chevalley Groups" by Steinberg.


Note above that in types $G\_2$, $F\_4$ and $E\_8$ the center of $Z(G\_{sc})$ is trivial so every irreducible representation is faithful. 


Also, for type $D\_{2l}$ the center is not cyclic so no irreducible representation is faithful.


