Let $\Phi$ be an irreducible root system, with positive roots $\Phi^+$ relative to the base $\Delta$. If $W$ is the Weyl group, how can I determine if $I$ belongs to $W$? Equivalently how can I see if the (unique) longest element in $W$ is $I$?

2$\begingroup$ see mathoverflow.net/questions/75231/… $\endgroup$ – Koen S Nov 25 '11 at 10:25
As Koen S points out, the longest element of an irreducible Weyl group is treated in an earlier question (in fact, it comes up in several questions). The question asked here presupposes a standard linear realization of the Weyl group, as occurs in the structure theory of a semisimple Lie algebra over $\mathbb{C}$ for instance. In this context, the standard answer given by Max is formulated as Exercise 5 at the end of Section 13 in my 1972 Springer graduate text. However one arrives at this conclusion, it obviously depends on the classification of irreducible root systems.
On the other hand, the question makes sense for any irreducible finite Coxeter group in its usual realization as a reflection group, and is approached in this spirit (via the Coxeter element) toward the end of Section 3.19 in my 1990 book on reflection groups and Coxeter groups.
P.S. For irreducible Weyl groups, a natural motivation for asking this question involves the criterion for all finite dimensional irreducible representations of the associated simple Lie algebra to be selfdual: If the given highest weight is $\lambda$, the dual has highest weight $w_0 \lambda$ (where $w_0$ is the longest element of $W$). This always coincides with $\lambda$ iff $1 \in W$.
Depends on what you mean with $I$...
If you mean an element $w\in W$ such that $w\Phi^+=\Phi^$, then this does always exist and is always the longest element $w_0$. See for example Section 1.8 in "Reflection Groups and Coxeter Groups" by Humphreys (who is also a regular here ;).
If you meant to ask whether $w\in W$ exists such that $w$ acts like $I$ on the natural geometric representation of $W$ (as defined in Section, then this is the case if and only if $W$ has nontrivial center, in which case the center has just two elements, $w_0$ and $1$, and you want $w=w_0$. Now obviously $w_0$ is in the center if and only if conjugation by $w_0$ is trivial. But in many cases, conjugation by $w_0$ induces a diagram automorphism. With some computations one can thus verify that $w_0$ is in the center for types $A_1$, $BC_n$, $D_{2n}$, $E_7$, $E_8$ and $F_4$. While you get a trivial center for type $A_n$ ($n>1$), $D_{2n+1}$ and $E_6$.
The mnemonic I use: if the diagram has a natural involution, then $w_0$ induces it, otherwise $w_0 = 1$. The only place this fails is in $D_n$, where one can switch the antlers, but shouldn't always, so how to remember when? In $D_4$, there's no natural involution, so that helps me remember that in $D_{2n}$, the involution is trivial.

7$\begingroup$ ...or one remembers that it depends on the parity of $D_n$ and uses the case of $D_3 = A_3$ and $D_2 = A_1 \times A_1$ to decide. $\endgroup$ – Geordie Williamson Nov 29 '11 at 13:19

1$\begingroup$ In the noncrystallographic cases it seems harder to come up with a useful mnemonic device, though the general criterion is always the same: 1 belongs to the irreducible finite Coxeter group iff all degrees of basic polynomial invariants are even. $\endgroup$ – Jim Humphreys Nov 29 '11 at 17:00

$\begingroup$ Also I is in the Weyl group if and only if the center of the simply connected group is a 2group $\endgroup$ – Jeffrey Adams Jan 25 '12 at 18:20