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 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 a torsion prime").

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 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").

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.

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 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 a torsion prime").