EDIT: I am still hoping for further responses to this question, but perhaps it isn't focussed enough and as I'm unsure what the best way is to improve it, I'll leave it as it is and let it sit here for the time being.

The following notion has appeared in my work and I am wondering if anyone can help me to understand it. I am not an algebraist and this seems like something that one might be able to say something about. See below for my motivation for this question. I put it there because the context of it isn't necessary to understand the question. Thank-you for any insight offered.


The definition of a module over a ring $R$ is an Abelian group $M$ along with a homomorphism $R \to EndM$ (the module is 'left' or 'right' depending on which multiplication we choose for the endomorphism ring). A homomorphism of $R$-modules is a homomorphism of Abelian groups $\varphi : M \to N$ that makes

$$\begin{array}[c]{ccccc} &&EndM&&\\ &\nearrow&&\stackrel{\circ\varphi}{\searrow}&\\ R&&&&Hom(M,N)\\ &\searrow&&\stackrel{\varphi \circ}{\nearrow}&\\ &&EndN&& \end{array}$$

commute, where the maps on the left are ring homomorphisms (defining the modules) and the maps on the right are given by pre- and post-composition, and where $Hom(M,N)$ is considered to be merely a set. I want to consider a different definition of homomorphism.


Our definition of ~homomorphism shall be a pair $(\phi,\varphi)$ where $\phi$ is an endomorphism of $R$ and $\varphi$ is a homomorphism of Abelian groups such that

$$\begin{array}[c]{ccccc} R&\rightarrow&EndM&&\\ &&&\stackrel{\circ\varphi}{\searrow}&\\ \small{\phi} \downarrow&&&&Hom(M,N)\\ &&&\stackrel{\varphi \circ}{\nearrow}&\\ R&\rightarrow&EndN&& \end{array}$$


Does this definition make sense, is it useful, and if so where can I read about these structures?

It is true that any such ~homomorphism induces a homomorphism (in the usual sense) from $M$ to a new module with ring action defined by $\phi$ and that of $N$. However, this new module depends on $\phi$; you can't do this for all ~homomorphisms at once. You'll see from my motivation that this is of no use to me.

I would also like to know the merits of such a definition applied to modules over different rings (replace the lower copy of $R$ in the diagram above with another ring $S$). Some interesting things seem to happen, for example the inclusion of a real vector space into its complexification is a ~homomorphism; more generally a complex vector space can have real subspaces in the genuine morphism sense. Complex conjugate-linear maps are also ~homomorphisms. One has to be careful though, because a bijective ~homomorphism is not necessarily an ~isomorphism. This happens in other categories too, e.g. a smooth bijection need not be a diffeomorphism etc., so I see no problem immediately.

Motivation from geometry (skip if you like)

My motivation for a different definition of homomorphism comes from geometry. The tangent space of a (locally-irreducible) hyper-Kähler manifold can be considered as a copy of the standard representation of $Sp_n$ on $\mathbb{R}^{4n}$. The action of $Sp_n$ preserves the subalgebra $\mathbb{H} \subset End\mathbb{R}^{4n}$ pointwise (where this time we mean linear endomorphisms). We can therefore consider the tangent space to be a left $\mathbb{H}$-module and the group $Sp_n$ to consist of module homomorphisms. Consider now the tangent space of a quaternionic-Kähler manifold. This can be considered as the representation of $Sp_n \cdot Sp_1$ on $\mathbb{R}^{4n}$. The group action preserves the same subalgebra $\mathbb{H} \subset End\mathbb{R}^{4n}$ but this time the induced action on this subalgebra is by $Sp_n \cdot Sp_1 \to SO_3$ (the second factor) and this is an irreducible action. In fact $SO_3$ is the automorphism group of the real algebra $\mathbb{H}$ (and also of the ring $\mathbb{H}$). In this situation we cannot consider the tangent space to be an $\mathbb{H}$-module, but rather as a module in the twisted sense above.


"I would also like to know the merits of such a definition applied to modules over different rings"

This is called "change of rings" -- Cartan, Eilenberg, Homological Algebra, Sec.2.6.

  • $\begingroup$ @ William Shears You are welcome! $\endgroup$ – Boris Novikov Jul 6 '13 at 6:22

This question is over 4 years, but I'll answer anyway since I just came across the exact same structure; check post here. Your construction makes sense, and "restriction of scalars" for modules over a ring is all you need to understand your twisted homomorphisms.

Any ring homomorphism $\phi:R\to S$ will "pullback" any $S$-module to an $R$-module. Namely, given a homomorphism $\rho:S\to \text{End} N$, you may compose it with $\phi$ $$ \text{Res}_\phi \rho:R\overset{\phi}{\rightarrow} S\overset{\rho}{\rightarrow}\text{End} N\, $$ to give $N$ an $R$-module structure. In particular, for any ring $R$ automorphism this construction induces an automorphism of its category of $R$-modules.

Now, a homomorphism of modules $M,N$ over different rings $R,S$, respectively, is a pair $(\phi, \varphi)$ such that the following diagram commutes: $$ \require{AMScd} \begin{CD} R\otimes M @> > > M\\ @V \phi\otimes \varphi V V @V V \varphi V\\ S\otimes N @> > \rho > N \end{CD} $$ Note that you get the usual $R$-module homomorphisms for $S=R$, $\phi=\text{id}_R$. On the other hand, this is exactly the same thing as the commuting diagram: $$ \require{AMScd} \begin{CD} R\otimes M @> > > M\\ @V \text{id}_R \otimes \varphi V V @V V \varphi V\\ R\otimes N @> > \text{Res}_\phi \rho > N \end{CD} $$ That is, a regular $R$-module homomorphism between $M$ and the restriction of $N$ along $\phi$; and these you can understand entirely within $R$-mod.

I see you want to understand $\text{Hom}_R(M, \text{Res}_\phi N)$ for fixed $M,N\in R$-mod and all $\phi\in\text{Aut}(R)$. But this is as hard as (or as easy as) understanding $\text{Hom}_R(M,N)$ since it sits inside our object by considering $\phi=\text{id}_R$. You also have a map to $\text{Aut}(R)$ by forgetting $\varphi$. So, this fits into a short exact sequence: $$ 1\to \text{Hom}_R(M,N)\hookrightarrow \text{Hom}_R(M, \text{Res}_\phi N)\to \text{Aut}(R)\to 1\,. $$

Hope this helps. Maybe, someone else could tune in and say a bit more. (Qiaochu's answer here might help you go a little further.)

  • 1
    $\begingroup$ Thank-you for your answer Carlos! This question is indeed so old that I am no longer a mathematician, but I read your answer with interest. $\endgroup$ – William Shears Apr 29 '18 at 21:11

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