A question on minuscule representations and $SL(2,\mathbb{C})$ representations Let $V$ be a minuscule representation of a complex semisimple Lie group $G$. After choosing a set of positive roots, and a corresponding set of simple roots of $G$.
edit 3: I use the following notation, $(-,-)$ denotes the standard ad-invariant inner product on a (fixed) Cartan subalgebra, while $<u,v> = (u,v)/(u,u)$, for $u\neq 0$ and $v$ elements in that Cartan subalgebra. I realize now that this notation is not standard, for which I apologize. 
Q1: is there a unique dominant weight of $V$, with respect to this choice of positive and simple roots?
Q2: assuming the answer to Q1 is yes, let $N = \sum <\alpha, \lambda>$, the sum being over all positive roots $\alpha$. Is $N$ an integer? (this will be the case if $\lambda$ is a so-called integral weight)
(Q1 and Q2 are standard material, that I can look up, but I am mostly interested in the next questions)
Q3 (edited following Prof. Webster's answer): Consider the principal homomorphism from $SL(2,\mathbb{C})$ in $G$. Is it true that the restriction (meaning pullback) of the minuscule representation to that $SL(2, \mathbb{C})$ always has an irreducible component which is isomorphic to $S^N(\mathbb{C}^2)$, where $N$ is as in Q2?
Note 1: I am new to representation theory, so I apologize if my questions are elementary to the experts.
Note 2: a paper by Prof. Benedict Gross deals with related topics, but he deals in that paper with the Langlands dual of $G$ (and I am not yet familiar with Langlands duality).
edit 1: I have fixed the definition of $N$ in Q2 (I had mistakenly written it initially in terms of the Weyl vector).
edit 2: I have edited Q3 in light of Prof. Webster's answer, and I have deleted Q4.
 A: The answers to Q1 and Q2 are both yes, by standard Lie theory.
(Note: This part of the answer is in response to an older version of the question, which asked if the restriction is irreducible.)  By some standard properties of nilpotent orbits, a representation will only have this property if it does for the principal $\mathfrak{sl}(2)$.  This is equivalent to the claim that the raising element $X$ in the $\mathfrak{sl}(2)$ has a single Jordan block, which is an open property, and regular nilpotents are dense in the set of nilpotents.  
Sometimes this happens, and sometimes it doesn't.  For example, if you consider the $k$th wedge power of the vector representation for $SL(n)$, then the highest weight for the principal $SL(2)$ is $n-1+(n-3)+\dots +(n-2k+1)$ and the dimension is $\binom nk$.  I don't think these ever match outside of very degenerate cases. For $n=4$ and $k=2$, the first interesting case, you get highest weight $4$ and dimension $6$, so there's an extra invariant under the principal $SL(2)$.  
EDIT:  Since the question has now changed, the answer to Q3 is (basically) yes, but this is nothing special about minuscule representations.  You just have to look at the highest weight vector and look at the weight it has under the action of the Cartan of the principal $\mathfrak{sl}(2)$.  This Cartan is spanned by the unique element $H$ of the Cartan of the whole Lie algebra with $\alpha(H)=2$ for all simple roots.  By the usual argument, we can see that this is $2\rho^\vee$, that is, the sum of the positive coroots.  So, the highest weight vector generated an $\mathfrak{sl}(2)$ representation with highest weight $2\rho^\vee(\lambda)$.  If your group isn't simply laced, this isn't quite the same as $\langle 2\rho, \lambda\rangle$ (which is where the Langlands duality comes in).  
