morphisms representable by algebraic spaces vs morphisms representable by schemes So I've been working with moduli stacks in algebraic geometry for a while now, with no formal training in the technicalities of the theory of algebraic stacks (ie, I've read a few articles and I learn what I need, without having spent much time doing exercises or working through examples/counterexamples).
One of the difficulties I've experienced is that while discussing morphisms of algebraic stacks, some authors only require certain morphisms to be representable by algebraic spaces, while others require them to be representable (by schemes).
Of course I've always found the latter to be more comprehensible (since it saved me from having to learn too much about alg. spaces on the way to working with alg. stacks). On the other hand, all the really comprehensive references on the subject usually fall into the former category.
For example, the stacks project's definition of an algebraic stack only requires the diagonal to be representable by alg spaces, while Gomez's article https://arxiv.org/pdf/math/9911199v1.pdf requires that it be representable (by schemes).
So far, whenever I read "representable by alg spaces", I sort of "pretend" that it says "representable by schemes" and proceed with that assumption.
My questions are:


*

*What is the difference between requiring the diagonal of an algebraic stack to be representable by spaces vs representable by schemes? When are the two notions equivalent? What pitfalls are there to avoid?

*In general, what are some good illustrative examples of morphisms of alg stacks which are only representable by spaces, but not by schemes? When are the two notions equivalent? What pitfalls are there to avoid?
 A: To answer question 2, the best example I know is $\mathscr{M}_1$, the stack of (proper smooth geom. connected) curves of genus 1. Indeed, Raynaud has contructed an elliptic curve $E\to S$ over a scheme $S$ and an $E$-torsor $X\to S$ which is (an algebraic space but) not a scheme.  
This implies two things. First, in order to define  $\mathscr{M}_1$ we are forced to take "curve" to mean "algebraic space in curves": if we insist that curves must be schemes, the resulting $\mathscr{M}_1$ will not be a stack for the flat topology, because the above $X$ is locally a (projective) scheme for the flat (even étale) topology on $S$.
Concerning the diagonal, put $I:={\underline{\mathrm {Isom}}}_{\mathscr{M}_1}(E,X)$. There is a monomorphism $X\to I$ (in fact, a closed immersion) identifying $X$ with the subsheaf of $E$-torsor isomorphisms. So, $I$ is not a scheme because $X$ isn't. Viewed as a morphism $S\to \mathscr{M}_1$,  $I$ is the pullback of the diagonal under the morphism $S\to \mathscr{M}_1\times\mathscr{M}_1$ given by $(E,X)$. Hence the diagonal is not representable in the scheme sense.
A: This may not directly answer your question. However, one reason to introduce the notion of an algebraic space is that sometimes it is easier to show that the diagonal is representable by an algebraic space than by a scheme.  Consider the following example. Let $G$ be a smooth algebraic group over a field $k$. Consider the classifying stack BG that parametrizes principal bundles that are locally trivial in the fpqc topology. Since $G$ is smooth over a point, and smooth surjective morphisms \'{e}tale locally admit a section, we note that these bundles are a fortiori trivial in the \'{e}tale topology.
Let $T$ be a scheme and suppose we have a morphism 
$$T \stackrel{(P_1, P_2)}{\longrightarrow} BG \times BG.$$
Here the $P_i$'s are two principal bundles over the scheme $T$. The fiber product
$$T \times_{BG \times BG} BG \cong \underline{\operatorname{Isom}}_T(P_1, P_2)$$
where the right hand side is the functor of isomorphisms between $P_1$ and $P_2$. Now choose an \'{e}tale cover $T' \to T$ that trivializes both the $P_i$'s (for instance if $T_i \to T$ trivializes $P_i$ then we can just take $T' := T_1 \times_{T} T_2$). Hence
$$\underline{\operatorname{Isom}}_T(P_1, P_2) \times_T T' \cong \underline{\operatorname{Isom}}_{T'}(G_{T'}, G_{T'}) \cong h_{G \times T'}.$$
This shows that the base change of the Isom functor by an \'{e}tale cover is a scheme, from which we conclude that Isom is an algebraic space. 
