Let $G$ be a group and $F:G^n \to G$ with the following property: If $x_1,…,x_n,h \in G$, then $F(hx_1,…,hx_n)=hF(x_1,…,x_n)$. Is there a name for this type of function property? It is something I’ve been investigating lately. For instance, if $G$ is a vector space and $F$ outputs the average vector, then $F$ has this property.
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If it is a vector space then what is $hx_1$? $h+x_1$? – markvs Dec 19 '21 at 05:25
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@markvs yes. ab is just the group operator applied to a and b. If we’re in a vector space, then this becomes vector addition. But this could also be, say, SO(n), and ab corresponds to the b rotation followed by the a rotation. – Spencer Kraisler Dec 19 '21 at 05:27
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4$G$-equivariant map with respect to the diagonal action on $G^n$. – tkf Dec 19 '21 at 05:30
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That does not seem to be useful for non-Abelian groups. I do not see why would anybody study such maps. – markvs Dec 19 '21 at 05:33
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2@markvs I assure you it’s very useful, especially for non-abelian groups. Basically means F cares only about the relative states of the n inputs, not their global states. – Spencer Kraisler Dec 19 '21 at 05:35
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1Although for $n=3$, people study various centers of triangles in the Cayley graph and the center $F(a,b,c)$ of triangle $abc$ usually satisfies this property. – markvs Dec 19 '21 at 05:36
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@SpencerKraisler: I do not need your assurances. A reference or two would be welcome. – markvs Dec 19 '21 at 05:37
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For example, https://arxiv.org/abs/1405.0757. – markvs Dec 19 '21 at 05:39
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Or this: https://people.maths.ox.ac.uk/drutu/rd-revised.pdf, or this: https://arxiv.org/pdf/0810.1969.pdf . – markvs Dec 19 '21 at 05:42
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1Given any $H:G^{n-1}\to G,$ $H$ determines a unique $F$ as $$F(g_1,\dots,g_n)=g_1H(g_1^{-1}g_2,g_1^{-1}g_3,\dots,g_1^{-1}g_n).$$ Also, this determines all such $F.$ – Thomas Andrews Dec 19 '21 at 06:00
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@markvs Apologies. One main concept is the karcher mean (https://hal.inria.fr/hal-00938320/document). It generalizes the Euclidean mean and has this property. I was wanting to study this particular property since it could lead to insight. – Spencer Kraisler Dec 19 '21 at 06:09
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@SpencerKraisler: For various centroids, the property you mention does not come alone. For example, $F(x_1,x_2,...x_n)$ and $F(x_{\sigma 1},...,x_{\sigma n})$ are usually equal where $\sigma$ is any permutation of $1,...,n$. I never heard about karcher mean, but it probably satisfies the same property. – markvs Dec 19 '21 at 06:25
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4One might call that a homogeneous function of degree $1$. A cursory search for an paper where $G$ is a specifically group didn't turn anything up, but the "Homogeneity under a monoid action" section of the linked article applies, since all groups are monoids. – Eric Towers Dec 19 '21 at 06:52
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@markvs for sure, and many more. But this particular property interests me honestly. I guess it's related to equivariant transformations and studying $G^N/G$. – Spencer Kraisler Dec 19 '21 at 19:02
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The set of such functions are in one-to-one correspondence with the set of all functions $G^{n-1}\to G.$
For example, if $H:G^{n-1}\to G$ we can define:
$$F(g_1,\dots,g_n)=g_nH(g_n^{-1}g_1,g_n^{-1}g_2,\dots,g_n^{-1}g_{n-1})$$
On the other hand, given an $F,$ we can get back $H$ by: $$H(g_1,\dots,g_{n-1})=F(g_1,\dots,g_{n-1},1).$$
So such functions $F$ don’t seem too interesting.
One might ask, more generally, if $G$ acts on a set $X,$ then $G$ acts on $X^n,$ and what can we say about functions $F:X^n\to X$ which is a map in the category of sets acted on by $G?$ In your case, $X=G.$
This might be more complicated. For example, if $G$ acts on $X$ $2$-transitively, and $n=2,$ then $F$ is entirely determined by one value of the form $F(x,x)$ and one value of the form $F(x,y), x\neq y.$ So there are at most $|X|^2$ such functions in that case.
In general, $X^n/G$ is very complicated, but when $X=G$ with the simple action, it is fairly simple.
Thomas Andrews
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My bad I should clarify. I don’t see how there’s a bijection. For every F, there’s an H. And for every H, there’s an F. But that doesn’t imply injectivity, right?. – Spencer Kraisler Dec 19 '21 at 06:40
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1Well, injectivity is easy to prove. If $H_1,H_2$ both give $F,$ then you can fairly easily show that $H_1=H_2$ by applying the definition for $F$ when $g_n=1.$ – Thomas Andrews Dec 19 '21 at 06:47
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I see, thank you. Can you elaborate what you mean by the second part of your answer? Also if you have any references for this type of stuff, equivariant functions, I'd highly appreciate it. – Spencer Kraisler Dec 19 '21 at 18:59
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1What part of the second section don’t you understand? Be specific. I can only help you understand specific stuff. If you don’t know about group actions, not sure comments here are a great place to teach and learn them. – Thomas Andrews Dec 19 '21 at 19:06
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actually that’s fine. You’ve answered a lot. I’ll be looking into the properties of G^n/G – Spencer Kraisler Dec 19 '21 at 19:23
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@SpencerKraisler To give an idea of the last part, Thomas Andrews is saying that in general $F:X^n\to X$ is determined by it's value on a representative of each orbit of the action by $G$ on $X^n$. ($X^n/G$ is the set of orbits). Thus there are at most $|X|^{|X^n/G|}$ such functions. In particular, it happens that $|G^n/G| = |G|^{n-1}$, and in the $2$-transitive case, $|X^2/G| \le 2$. – Milten Dec 19 '21 at 20:20
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This bijection is the standard identification of the homogenous and inhomogeneous bar resolutions of a group. See for example https://math.stackexchange.com/questions/65531/what-is-the-motivation-for-defining-both-homogeneous-and-inhomogeneous-cochains – tkf Dec 20 '21 at 18:37
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I would like to generalise Thomas Andrews's answer a bit more. Let $G$ act on two sets $X$ and $Y$, and consider functions $F:X\to Y$ such that $F(gx) = gF(x)$. In the original setting, we have $X=G^n$, $Y=G$. As Thomas Andrews has pointed out, $F$ is determined by its value on a representative of each orbit in $X/G$. This is of course because if $F(x_0)$ is known, then $F(gx_0)=gF(x_0)$ is forced for all $gx_0$ in the orbit of $x_0$.
In the original setting, Thomas Andrews showed (in slightly different words) that the functions $F$ are in in fact in a one-to-one correspondence with the set of all functions $G^n/G\to G$ (i.e. functions $X/G\to Y$). When do we have this nice property in general?
For each orbit in $a\in X/G$, choose a canonical representative $\psi(a)\in a$. Let's write $\hat x = \psi([x])$ for the representative of $x$'s orbit. Now we want to define $F$ by choosing a value at each orbit representative, and then setting $$ F(x) = F(g_x \hat x) := g_x F(\hat x). $$ This isn't well-defined in general, because there might be several choices for $g_x$. But if the action on $X$ is free, then $g_x$ is unique by assumption, and we're golden. Thus,
If $G$ acts on $Y$ and acts freely on $X$, then the functions $F:X\to Y$ such that $F(gx)=gF(x)$ are in a one-to-one correpondence withthe set of all functions $X/G \to Y$.
We recover the corresponding function $E:X/G \to Y$ by simply $$ E(a) = F(\psi(a)). $$ Note however that the concrete correspondence is not unique or natural, since it depends on our arbitrary choices of representatives when defining $\psi$.
Lastly, take the special case $G\le H$, where $G$ is a subgroup of $X=H$ and acts by left mulitplication. Then $h=g_h\hat h \implies g_h = h(\hat h)^{-1}$, so the action is free ($g_h$ is unique), and our result applies. This was the case in the original setting, if we identify $G$ with the diagonal subgroup of $G^n$. In this case $$ \widehat {(g_1, \ldots, g_n)} := (g_n^{-1}g_1, \ldots, g_n^{-1}g_{n-1}, 1). $$
Milten
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You can say the group operation is left-distributive over $F$.
When writing about this, make sure you include your definition, so it's clear for everyone.
Paul
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What definition are you talking about? I felt I defined everything necessary, no? – Spencer Kraisler Dec 19 '21 at 19:00
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3@SpencerKraisler I think the point is, if you use the term “left-distributive” somewhere, define it there first, because it is not a common or agreed-upon term. Not that you didn’t define it in your question. – Thomas Andrews Dec 19 '21 at 19:31
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