Are the weak topologies of $\ell_1$ and $L_1$ homeomorphic?
Strangely may it sound, the question seeks contrasts between norm and weak topologies of Banach spaces from the non-linear point of view. And indeed, no linear homomorphism between weak topologies of non-isomorphic spaces exists as prevented by the Closed Graph Theorem. Note that the fact that $\ell_1$ is a Schur space yet $L_1$ is not is a priori not helpful, unless I am missing something trivial.
However, all infinite-dimensional, separable Banach spaces are homeomorphic. For weak topologies this is not so, as the weak topology of a reflexive space is $\sigma$-compact whereas this does not hold for $L_1$, say. We could then ask:
Is there a pair of non-isomorphic, non-reflexive Banach spaces whose weak topologies are homeomorphic?
31.03.2022: As noticed by Jerzy, the weak topologies of $\ell_1$ and $C[0,1]$ are not homeomorphic.
Best Answer
The weak topologies of the spaces $\ell_1$ and $L_1$ are not homeomorphic because of the following
Theorem. Assume that $X,Y$ are two Banach spaces whose weak topologies are homeomorphic. If $X$ has Shur's property, then $Y$ has Shur's property, too.
Proof. For a topological space $T$ let $T_s$ be the sequential coreflection of $T$, which is $T$ endowed with the topology consisting of sequentially open sets. A subset $U\subseteq T$ is sequentially open if for any convergent sequence $(x_n)_{n\in\omega}$ in $T$ with $\lim x_n\in U$ there exists $m\in\omega$ such that $x_n\in U$ for all $n\ge m$.
Assume that $h:X_w\to Y_w$ is a homeomorphism between the Banach spaces $X,Y$ endowed with the weak topologies. The homeomorphism $h$ remains a homeomorphism betweem the sequential coreflections $X_s$ and $Y_s$ of the spaces $X_w$ and $Y_w$, respectively. Since $X$ has Shur's property, the sequential coreflection $X_s$ of $X_w$ coincides with $X$. Since $X=X_s$ is a Baire space, so is the space $Y_s$. The Baireness of the space $Y_s$ implies that the closed unit ball $B_Y$ of $Y$ has non-empty interior in $Y_s$. Let $y_*$ be an interior point of the set $B_Y$ in $Y$. Assuming that $Y$ fails to have Shur's property, we can find a null sequence $(y_n)_{n\in \omega}$ in $Y_w$ consisting of vectors of norm $\|y_n\|=3$. Then $(y_n+y_*)_{n\in\omega}$ is a sequence in $Y_s\setminus B$ that converges to $y_*$, which is not possible as $y_*$ is an interior point of $B$ in $Y_s$. This contradiction implies that the Banach space $Y$ has Shur's property.
The proof of the above theorem suggests the following characterization of the Shur property.
Characterization. A Banach space $X$ has Shur's property if and only if the sequential coreflection $X_s$ of the weak topology of $X$ is a Baire space if and only if $X_s=X$.
Remark. Concerning the second question, it can be shown that any reflexive separable infinite-dimensional Banach spaces endowed with the weak topologies are sequentially homeomorhic (= the sequential coreflections of their weak topologies are homeomorphic), see this paper for more information in this direction.