Suppose $G$ is a connected reductive group over an algebraically closed field. Then given a maximal torus $T$, we can define a Weyl group $W$ and consider $T^W$, the Weyl-invariants of $T$. This clearly contains the center $Z(G)$ but can be larger. For instance, in $\mathrm{PGL}\_2$, the element with $\mathrm{GL}\_2$ representative $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ is not central but is a Weyl-invariant.


Are the connected components of these groups the same? Is there a simple condition one can place on $G$ that forces these groups to be equal? 



Fixed Points of the Weyl Group action on a Maximal Torus and the Center of a Reductive Group
Notice that nothing in the problem is harmed by base change, so that it doesn't matter that the ground field is algebraically closed.


The identity component of $T^W$ is generated by the images of the $W$-invariant cocharacters of $T$, and the $W$-invariant part of the cocharacter lattice consists precisely of the central cocharacters. (Proof: try applying the reflection in a root to a cocharacter.) Thus we always have the desired equality of identity components.


In general $X^\*(T^W) = X^\*(T)\_W$, whereas $X^\*(\mathrm Z(G)) = X^\*(T)/Q$, where $Q$ is the root lattice; so your question about equality $T^W = \mathrm Z(G)$ is asking whether
$$
Q = X^\*(T)(W) = \mathbb Z\{\langle X^\*(T), \alpha^\vee\rangle\alpha \mid \alpha \in \Phi\}.
$$
This certainly happens if $G$ is simply connected (since then the fundamental weights lie in $X^\*(T)$), but not only then (for example, it also works if $G = T$ is a torus). I think that it is probably equivalent to the derived group of $G$ being simply connected, but I have not checked (and now [@KasperAndersen](https://mathoverflow.net/a/307897) points out that it is not true! Let's guess again: "the fundamental group of $\mathcal D(G)$ has odd order" might do, or perhaps "2 is not a torsion prime").


