why does a fundamental representation of SL(n) have only one dominant weight? I am asking this because /\i(v) is a fundamental representation , but that can happen only if theres only one dominant weight and all others are conjugates under the weyl group action to it.
[Math] dominant weights
rt.representation-theory
Related Solutions
For type A, a combinatorial interpretation of the entries of the inverse matrix was given by O. Egecioglu and J.B. Remmel, A combinatorial interpretation of the inverse Kostka matrix, Linear Multilinear Algebra 26 (1990) 59-84. The formula involves a lot of cancellation which suggests that the entries are relatively small. Another interpretation of the entries was given by H. Duan, On the inverse Kostka matrix, J. Combinatorial Theory (A) 103 (2003), 363-376.
Your question is really about virtual vector spaces: what is a virtual vector space?
Once you know what a virtual vector space is, then there is only a small step to the answer of your question.
There are a few possible answers:
1• A virtual vector space of a pair of vector spaces. Equivalently, it's a $\mathbb Z/2$-graded vector space. You should think of the pair $(V,W)$ as being the formal difference $V-W$. This approach has the downside that it makes it unclear what an isomorphism between virtual vector spaces should be.
2• A vector space of dimension $n$ is the same thing as a point (sic!) of the topological space $BU(n)$. A virtual vector space is then a point of the space $BU:=\mathrm{colim}_{n\to \infty} BU(n)$. This approach is not geometric at all, but works very well for talking about virtual vector bundles on a space $X$: these are continuous maps $X\to BU$.
3• Fix an infinite dimensional vector space $U$ and a polarization $U=U_-\oplus U_+$ (both $U_-$ and $U_+$ are infinite dimensional). A virtual vector space is a (necessarily infinite dimensional) subspace $V\subset U$ such that $V\cap U_-$ is of finite codimension inside $V+U_-$.
4• Fix an infinite dimensional Hilbert space $H$. A virtual vector space is a Fredholm operator $F:H\to H$ (i.e., an operator with finite dimensional kernel and cokernel). This is related to definition 1 by assigning to the Fredholm operator $F$ the pair $(\ker(F),\mathrm{coker}(F))$.
5• A virtual vector space is an object of the bounded derived category of vector spaces: i.e., it's a chain complex.
All these definitions (with the exception of 2, for which it's more involved) can be easily adapted to the context of $G$-representation, you just need to replace "infinite dimensional vector space" with "$G$-rep that contains each irrep infinitely often".
Best Answer
The question is not well-formulated, as Bugs Bunny observes. In the case of the special linear group or Lie algebra (say over the field $\mathbb{C}$), it happens to be true that the fundamental dominant weights are precisely the minuscule ones (defined by the condition that the weight is nonzero while all pairings with arbitrary positive coroots are 0 or 1), equivalent to being minimal (and nonzero) in the usual partial ordering of dominant weights. In particular, for an irreducible representation having such a highest weight all its weights are conjugate under the Weyl group, with only the highest weight therefore being dominant.
For irreducible root systems of other than type $A$, the minuscule weights are also fundamental but there are very few of them. For example, the 0 weight always occurs in irreducible representations for type $G_2$ and rules out existence of minuscule weights. Bourbaki is the most standard reference (Groupes et algebres de Lie, VIII, 7.3), but in textbooks the details are usually left as an exercise.
In type A, it's a standard exercise to compute the highest weights occurring in the various exterior powers of the standard representation (whose highest weight is the first fundamental weight) and see directly that these are minuscule and in particular fundamental.