60

I'd like to think that I understand symmetry groups. I know what the elements of a symmetry group are - they are transformations that preserve an object or its relevant features - and I know what the group operation is - composition of transformations. Given a polyhedron or wallpaper tiling or whatever, I could probably start spotting the symmetries, which would entail listing out elements of the symmetry group, and then I could start filling in the multiplication table.

Penrose attaches a group to impossible figures to capture their inherent ambiguity, and I'd like to grok these groups like I do symmetry groups. Take a prototypical example, the tribar:

penrose tribar

He names the "ambiguity group" $G=\Bbb R^+$ (positive numbers under multiplication) to describe possible distances of points. We can split the figure up into three components, as above, and interpret them as being disconnected from each other in three-space but from our perspective they seem to make a single figure. For convenience, I think we should let $A_{ij}$ denote points on the figures as well as represent their distances from the origin, interchangeably.

One can define the relative distances by $d_{ij}=A_{ij}/A_{ji}$. Since $d_{ji}=d_{ij}^{-1}$, there are only three relevant proportions: $d_{12}$, $d_{23}$, and $d_{31}$. According to Penrose, the $d_{ij}$s do not actually depend on our choice of overlapping points $A_{ij}$, but this seems wrong to me: varying the points $A_{ij}$ through the overlap regions will change them linearly and so any ratio $d_{ij}$ will only remain invariant if $d_{ij}=1$ to begin with. But probably this quibble is unimportant.

One can scale the distances the components $Q_1,Q_2,Q_3$ are from the origin without affecting our perception of them. (Perhaps consider our "perception" of them to be their radial projection onto the unit sphere, or something.) The effect of scaling one of these $Q_i$ by a factor of $\lambda$ on the $d_{12},d_{23},d_{31}$ is to scale one of them by $\lambda$, a second by $\lambda^{-1}$, and leave the third unchanged.

If the $Q_1,Q_2,Q_3$ were compatible and could be combined into a single figure, then such a configuration would have $(d_{12},d_{23},d_{31})=(1,1,1)$. If they were compatible but the components were separated by independent scalings $q_1,q_2,q_3$ (respectively) then we'd have

$$(d_{12},d_{23},d_{31})=\left(\frac{q_1}{q_2},\frac{q_2}{q_3},\frac{q_3}{q_1}\right). \tag{1}$$

Note that $\tau=d_{12}d_{23}d_{31}$ is an invariant, in the sense that scaling the components independently does not change the value of $\tau$. The compatibility situation $(1)$ occurs precisely when $\tau=1$.

Penrose defines the group $H$ to be the tuples $(d_{12},d_{23},d_{31})\in(\Bbb R^+)^3$ modulo the rescalings by $\lambda$ and modulo the elements of the form $(1)$. As I understand it, the invariant $(d_{12},d_{23},d_{31})\mapsto\tau$ is a bijection $H\to\Bbb R^+$. But now here are my questions.

$\hskip 1.4in$ necker cubes

$\sf \color{Fuchsia}{(A)}$ How do we know what the ambiguity group is? The tribar's ambiguity group is $G=\Bbb R^+$. With the Necker cubes above, Penrose says the ambiguity group is $G=\Bbb Z_2$. Is the ambiguity group meant to parametrize the possible positions of the individual pieces of the figure? Where does the group operation of $G$ actually come into play?

$\sf \color{Fuchsia}{(B)}$ How do we know what pieces to cut a figure up into? Since the tribar has obvious threefold symmetry, that kind of inspires the choice of three pieces. But it seems that with the congruence relation used to define $H$, we could choose to put any two of those pieces together into one component and fix its position, only letting the last component vary (which would be one degree of freedom, exactly as $H\cong\Bbb R^+$ predicts). So we could have cut into two pieces. Or we ould cut into six pieces, or any number of pieces. Will it never matter how many pieces we choose? Why would we break apart the tribar's corners but not break apart the faces of the Necker cubes? What figures would we cut into pieces, and what figures would we do something else to? And with the latter figures, what would we do to find their $H$ group?

$\sf \color{Fuchsia}{(C)}$ What are the group elements and what is the group operation? As I mentioned with symmetry groups, it's intuitive what their elements are and what the operation is. But what about with $H$? It seems the elements are physically realizable configurations consistent with our perception, modulo altering the configuration in a way that wouldn't change our perception. The identity element would be the configurations in which our perception is actually correct and sensible. And the group operation seems to be ... I don't know. Presumably we could use componentwise multiplication of the representative tuples $(d_{12},d_{23},d_{31})$, or equivalently multiplication of the invariants $\tau$, which would make $H\to\Bbb R^+$ a group isomorphism, but how would this operation be meaningful or relevant?

$\sf \color{Fuchsia}{(D)}$ Why is this called a cohomology group? Yes, our $H^1(Q,G)$s are being called cohomology groups. I deliberately put off using that word as long as possible. (And so you've read this far. Suckers.) But in what sense are these groups cohomological? Are there higher cohomology groups $H^n(Q,G)$ and coboundary operators? Is this cohomology dual to some kind of homology of impossible figures? Probably I will be unable to understand answers to this question, as I don't really know what cohomology is in the first place. No time like the present?

Ultimately, I'd like to be able to look at an impossible figure and systematically derive its cohomology group, just like I can derive a figure's symmetry group. Or alternately, create impossible figures with given cohomology group. But perhaps the analogy isn't tenable, as cohomology groups aren't really symmetry groups at all.

Source: On the Cohomology of Impossible Figures.

whacka
  • 15,195
  • 25
  • 45

1 Answers1

36

To answer $\sf \color{Fuchsia}{(A)}$, one has to understand the meaning of the ambiguity group clearly first - indeed, your interpretation of the group is not entirely correct.

The punchline here is that the ambiguity group measures ambiguity, not quite impossibility (which is the job of the cohomology group defined later), of the given figure. For example, consider the tribar, and piece it up into $Q_1, Q_2, Q_3$, as done by Penrose. If you stretch $Q_1$ by a factor of $\lambda > 0$, shrink $Q_2$ by a factor of $\lambda$ (that is to say, stretch by $\lambda^{-1}$), and leave $Q_3$ as it is, the resulting object is really the same tribar, which is what Penrose was trying to convey through the following picture from the paper.

enter image description here

That is to say, modifying $Q_i$'s in this way gives you a symmetry of $X$. And you have such a symmetry for each $\lambda > 0$. Of course, collection of such symmetries $G$ is a group: consider one symmetry which stretches $Q_1$ by $\lambda$, shrinks $Q_2$ by $\lambda$ and leaves $Q_3$ as is and another one which does the same thing with $\lambda'$ this time instead. You can of course compose them, and the resulting symmetry stretches $Q_1$ by $\lambda \lambda'$, shrinks $Q_2$ by $\lambda \lambda'$, and leaves $Q_3$ invariant. This is the ambiguity group. This partially answers $\sf \color{Fuchsia}{(C)}$. Note that the origin of the name comes from the fact that the tribar is ambiguous in the sense that you can have different $Q_1, Q_2, Q_3$'s matching up to produce the same tribar. You can never tell the length of a bar in the tribar - it'll depend on the point of view you're seeing it from. And the "group of symmetries" $G$ captures all the ambiguity there is.

It is now clear that there is a bijection $G \to \Bbb R^+$, sending a symmetry to the corresponding $\lambda > 0$ in $\Bbb R^+$ (group under multiplication), which is in fact a homomorphism. Thus, $G \cong \Bbb R^+$.

That said, given the Necker cubes, the ambiguity is in the vertices. Pick any vertex $v$ from the figure. There are two point of views here - $v$ can either point upwards, or point downwards. If $v$ points upwards, then so does all the other vertices, and similar for the other choice. So two point of views, two choices - hence, the ambiguity group is $\Bbb Z/2\Bbb Z$. More formally, the symmetry is obtained from reflecting the whole figure (imagined as sitting in $\Bbb R^3$) along an appropriate hyperplane. Obviously, the symmetry group then is the cyclic group of order 2.

To answer $\sf \color{Fuchsia}{(D)}$. Yes, this is indeed a variant of Čech cohomology.

First, let me explain what cohomology really is. If $X$ is a sufficiently nice topological space, you can always set up a homeomorphism $X \cong T$ of $X$ with a simplicial complex $T$. Such a homeomorphism is called a triangulation of $X$. Note that there need not be a unique such homeomorphism, so there may be lots of ways to triangulate a given nice topological space.

Declare $\Delta_n(T)$ to be the free abelian group generated by $n$-simplices of $T$. Let $\Delta^n(T)$ be $\hom(\Delta_n(T), \Bbb Z)$, the dual of $\Delta_n(T)$. Elements of $\Delta^n(T)$ are called $n$-cochains. There is a natural map $\partial : \Delta^{n-1}(T) \to \Delta^n(T)$ taking an $n$-cochain $\psi : \Delta_{n-1}(T) \to \Bbb Z$ to the $n$-cochain $\psi \circ \text{bd} : \Delta^n(T) \to \Delta^{n-1}(T) \to \Bbb Z$ where $\text{bd}$ sends an $n$-simplex in $T$ to sum of signed faces of that $n$-simplex (signs chosen appropriately to take care of orientation). $\partial$ is called the coboundary map. One can check that the map $\partial \partial : \Delta^{n-1}(T) \to \Delta^n(T) \to \Delta^{n+1}(T)$ is the zero map. The sequence of maps

$$\cdots \to \Delta^{n-1}(T) \to \Delta^n(T) \to \Delta^{n+1}(T) \to \cdots$$

is then a cochain complex, that is, going two steps in the sequences lands you into $0$. As a consequence of $\partial \partial = 0$, we have that $\text{im} \partial \subset \ker \partial$. Elements of $\text{im} \partial$ are called coboundaries, and elements of $\ker \partial$ are called cocycles. As everything is abelian, $\text{im} \partial$ is also normal in $\ker \partial$. We set $H^n(X; \Bbb Z) := \ker \partial/\text{im} \partial$. This is called the simplicial cohomology of $X$, and is indeed independent of the triangulation $T$ we chose. It's a topological invariant of $X$. We'll see something similar happening in Penrose's paper.

If $X$ is the given ambiguous figure, let $\{U_i\}$ be a good cover of $X$. That is, $X = \bigcup U_i$ and each of $U_i$, $U_i \cap U_j$, $U_i \cap U_j \cap U_k$ and further intersections, are homeomorphic to balls $B^n$ (a more general condition would be that they are all contractible). Let each intersection $U_{i_1} \cap \cdots \cap U_{i_n}$ be denoted as $U_{i_1 \cdots i_n}$. Treat these sets as symbols (i.e., formal intersection), and distinguish $U_{ij}$ from $U_{ji}$ even though they are the same set.

$G$ be the ambiguity group of $X$. Define an $n$-dimensional cochain to be a function $\varphi^n : \{U_{i_1 \cdots i_n}\} \to G$ such that $\varphi^n(U_{i_1 \cdots i_k \cdots i_l \cdots i_n}) = \varphi^n(U_{i_1 \cdots i_l \cdots i_k \cdots i_n})^{-1}$ holds for any $k, l$, where $\{U_{i_1\cdots i_n}\}$ is the collection of all possible intersection of $n+1$ many sets from the cover. When $n=0$, it's just an assignment of a an element of $G$ to each $U_i$.

Let $C_n(X)$ denote the group of all $n$-dimensional cochains, where we multiply two cochains by multiplying their values in $G$. Define the coboundary map $\partial : C_{n-1}(X) \to C_{n}(X)$ given by sending $\varphi^{n-1}$ to the $n$-cochain $\varphi^n$ assigning to each $U_{i_i \cdots i_n}$ the number $\prod_j \varphi^{n-1}(U_{i_1 \cdots \hat{i_j} \cdots i_n})^{\varepsilon_j}$ where $\varepsilon_j = (-1)^n$ if $j$ is even and $(-1)^{n+1}$ if $j$ is odd, and $\hat{i_j}$ means that term is missing from the indices. So for example, in the tribar example, assigning to each $Q_i$ the number $q_i$ is a $0$-cochain and applying the coboundary map one gets an assignment of each (oriented) intersection $Q_{ij} = Q_i \cap Q_j$ to the number $q_i/q_j$, a $1$-cochain in our language.

The final thing to check is that $\partial \partial : C_{n-1}(X) \to C_{n}(X) \to C_{n+1}(X)$ is the zero map. This is a rather tedious and technical thing to check, but if the readers try verifying it, they will understand the motivation behind the rather bizarre definition of $\partial$ above. To sum up, we have that

$$\cdots \to C_{n-1}(X) \stackrel{\partial}{\to} C_n(X) \stackrel{\partial}{\to} C_{n+1}(X) \to \cdots$$

is a cochain complex. We can thus take it's cohomology $H^n(X; G) := \ker \partial/\text{im}\partial$ (the $G$ is there to indicate that the cochains are $G$-valued. Indeed, we could have put any group to be the coefficient group there, but it seems Penrose finds it natural to use the ambiguity group of $X$). For $n = 1$, this is precisely the $1$st cohomology group of Penrose.

Let us compute the cohomology group of the tribar. $H^1(X;\Bbb R^+)$ is the group of $1$-cocycles modulo $1$-coboundaries. As there are no nontrivial $2$-cochains, every $1$-cochain is a cocycle. That is to say, we're computing $1$st cohomology of the two-term cochain $$0 \to C_0(X) \stackrel{\partial}{\to} C_1(X) \to 0$$ which is just $C_1(X)/\text{im} \partial$. Note that this has a slight similarity with your interpretation (the definition of $H$), but is not quite the same. In any case, I can find a $1$-cocycle in $C_1(X)$ which is not a coboundary using the clever technique devised by Penrose: the $1$-cochain $\psi^1$ sending $Q_{ij} = Q_i \cap Q_j$ to $d_{ij} = A_i/A_j$ is independent of the choice of $A_k$'s, and I claim that it is a nontrivial cochain (i.e., is not a boundary). Well, if it was a boundary, then the equality $d_{ij} = \psi^1(Q_{ij}) = \psi^0(Q_i)/\psi^0(Q_j)$ would hold for some $0$-cochain $\psi^0$. But then we'd have $\tau(\psi^1) = d_{12}d_{23}d_{31} = 1$.

However, if we choose the placement of tribar appropriately so that $A_{12}$ and $A_{21}$ has the same distance from origin (i.e., $d_{12} = 1$) and $A_{23}$ is farther away from origin than $A_{32}$ (so that $d_{23} > 1$, then $A_{13}$ would automatically be closer than $A_{31}$ (that is, $d_{31} = d_{13}^{-1} > 1$), which implies $\tau(\psi^1)$ is strictly greater than $1$, forcing $\psi^1$ to be nontrivial. Once we have this nontrivial cochain, we can get $|\Bbb R^+|$-many nontrivial cochains by multiplying $\psi^1(Q_{12})$ by $1$, $\psi^1(Q_{23})$ by $\lambda$, and $\psi^1(Q_{31})$ by $1/\lambda$ for each $\lambda > 0$. This gives a natural identification with $G$. I think with some work you can show that these are all the nontrivial $1$-cocycles there are, and conclude $H^1(X; \Bbb R^+) \cong \Bbb R^+$. In general, it need not be true that $H^1(Q; G)$ is the same as the ambiguity group $G$, but the $1$-st cohomology group $H^1(Q; G)$ would be a module over the ambiguity group $G$.

I picked the open cover $\{U_i\}$ of $X$ to be good, i.e., all sets and all intersections are contractible. The reason is that by the nerve theorem, $X$ would have the homotopy type of the nerve $\mathcal{N}$ of $\{U_i\}$. Čech cohomology just computes cohomology of the nerve $\mathcal{N}$, but since this is homotopy equivalent to $X$, that is the same as taking cohomology of $X$, so it's independent of whatever good cover we chose. I think something similar must be at work for these ambiguous/impossible figures. This probably answers $\sf \color{Fuchsia}{(B)}$, but only Penrose knows what he had in mind.

One more thing I haven't addressed yet.

Ultimately, I'd like to be able to look at an impossible figure and systematically derive its cohomology group, just like I can derive a figure's symmetry group. Or alternately, create impossible figures with given cohomology group. But perhaps the analogy isn't tenable, as cohomology groups aren't really symmetry groups at all.

The first part of this answers gives a rough overview of how you could recognize the ambiguity group from a given ambiguous picture, so I presmume I have answered that bit. I do not know what a rigorous definition of the ambiguity group might be, but I presume it's some sort of inherent symmetry group of the chosen cover $\{U_i\}$. Also, certainly the analogy is tenable: if you read my answer to $\sf \color{Fuchsia}{(A)}$, you'd see that this ambiguity group is indeed some sort of symmetry groups which captures symmetries "induced from ambiguity". However, I do not know if you can construct an ambiguous figure such that the ambiguity group $G$ is isomorphic to a given group. Here's some food for thought: Assume $G$ is a finitely presented abelian group. Pick generators $x_1, \cdots, x_n$ of $G$. $F$ be the free abelian group generated by these. If $Y$ is some ambiguous figure with infinite cycle ambiguity group, then disjoint union of $n$ copies of $Y$ gives you a figure with ambiguity group isomorphic to $F$. Now given relators $r_i = 1$ for $G$, try identifying the copies of $Y$ appropriately to kill the symmetries corresponding to $r_i$. Can this be done?

PS: I wanted to add an explicit computation of the Penrose cohomology $H^1(X; \Bbb Z/2)$ of the Necker cube as a summary, but I'd need some pictures for that and I'm too ashamed of my drawing skills to add it. Maybe if OP (and others reading this) are really interested.

Balarka Sen
  • 13,020
  • 1
  • 41
  • 71
  • 1
    Very interesting answer. Please post. – Justpassingby Dec 30 '15 at 22:35
  • Yeah, it would be very interesting to add that part. – Hakim Dec 31 '15 at 22:27
  • One thing that slightly bothers me is that you seem to be saying that the impossibility of the tribar is represented by a nontrivial cocycle in the first cohomology group of the tribar, but the tribar itself isn't a well-defined topological space. The obstruction really lies in the first cohomology group of the (flat, 2-dimensional) annulus which contains the picture of the tribar (this is also what Penrose mentions, if I recall correctly). – Alex Provost Jan 09 '16 at 22:40
  • If my comment above is correct, what would be the appropriate 2-dimensional space to use that contains the picture of the Necker cubes, and what open cover should we use? (I.e., in what cohomology groups would the obstruction lie in this case?) – Alex Provost Jan 09 '16 at 22:41
  • @A.P. The obstruction lies in the nerve of the cover of the tribar we are using, which is a well defined abstract simplicial complex. I have noticed Penrose mention the annulus too, but I am not sure how it is relevant (note that I am just writing up _my_ interpretation of Penrose's paper, but it might as well be possible that this is not the case). – Balarka Sen Jan 09 '16 at 22:48
  • For the Necker cube, you need to look at the sub-picture consisting of three of the cubes with each face of the three cubes making another cube. Then pick open covers to be appropriate thickening of three of the cubes (or so I think it is - I figured it out when I seriously thought about it, but have forgotten the details). The nontrivial cocycle comes from choosing vertices of the cubes to point inwards/outwards. I guess I should add a picture once I get free. – Balarka Sen Jan 09 '16 at 22:52
  • I think I understand what you mean, and I can see how both approaches really amount to the same thing; but what really bothers me when I try to apply a similar reasoning to the Necker cubes. It feels like, no matter how I look at it, the corresponding space (either 2- or 3-dimensional) should be contractible, and so all the nonzero cohomology groups should vanish. How, then, can we understand the impossibility of the Necker cubes cohomologically? – Alex Provost Jan 09 '16 at 22:52
  • @A.P. Well, contractibility actually doesn't matter much here. Note that the coboundary map (defined rigorously above) here measures the perceptional ambiguity/impossibility of a given cocycle. So I'd say it's just the perception what matters. This is not quite the classical Cech cohomology. – Balarka Sen Jan 09 '16 at 23:00
  • I agree that this is not quite the usual setting for Cech cohomology, but in the end the coboundary map you defined is precisely the Cech coboundary map, and the obstruction lies in the Cech cohomology group of the annulus (or the simplicial cohomology of the nerve). – Alex Provost Jan 09 '16 at 23:31
  • In [this essay](http://www.ams.org/samplings/feature-column/fc-2014-10), the author finds that the impossibility of a Necker cube-esque figure (a part of a Penrose tiling, but which – like the tribar – exhibits circular symmetry) also lies in this group (with $\mathbb{Z}/2$ coefficients), but has to delete a point in the center in order to get an annulus-like shape again. – Alex Provost Jan 09 '16 at 23:31
  • @A.P. I am not sure if I see why the coboundary map defined is the Cech coboundary map. Can you elaborate on that? Also, in the essay, it seems to me that the center is deleted just to not get a 3-simplex in the nerve. Admittedly I haven't thought whether the involvement of the 3-simplex makes 2-cycles trivial. Interesting point nonetheless. – Balarka Sen Jan 09 '16 at 23:40
  • Nevermind, of course it is the Cech coboundary map. Just in the multiplicative sense. I misunderstood you. – Balarka Sen Jan 09 '16 at 23:42
  • 1
    I also assumed at first that the point is deleted in order not to have to deal with triple intersections, but if I try not deleting the point and considering three slightly larger open sets which intersect 3-fold, I can't even make sense of what the 2-cochain on the central set is supposed to represent in this setting. (According to my understanding, a 0-cochain represents a choice of perception, and a 1-cochain represents the degree to which two choices of perception agree.) And then there is still the problem of having trivial nonzero cohomology groups when dealing with a contractible space. – Alex Provost Jan 09 '16 at 23:52
  • @A.P. These are good points. Thanks for the comments (+1). I'll have to think through this tomorrow morning. – Balarka Sen Jan 09 '16 at 23:54
  • Thanks! I'm glad to finally have the opportunity to share these thoughts with someone else :-) – Alex Provost Jan 09 '16 at 23:56
  • In fact, I think it's safe to say that the Necker cubes in OP's picture are *not* impossible (this agrees with the trivial cohomology groups idea). They are *ambiguous*, because there is a $\mathbb{Z}/2$ action that reverses the perception of the vertices, but the same can be said of any of the three separate rods of the tribar (where in that case the ambiguity lies in their perceived distance), which aren't impossible by themselves. The Necker cubes can be assigned a *global* choice of perception which is coherent. – Alex Provost Jan 10 '16 at 13:55
  • @A.P. Three separate pieces of the tribar are not ambiguous. They can be coherently realized unambiguously as objects in $\Bbb R^3$. – Balarka Sen Jan 10 '16 at 14:50
  • But their distance with respect to the observer's eye is ambiguous. They can be realized in $\mathbb{R}^3$, with the additional $\mathbb{R}_{> 0}$-ambiguity when they are projected to a plane. And the Necker cubes can also be realized as objects in $\mathbb{R}^3$ (they're just a bunch of cubes), with the additional $\mathbb{Z}/2$-ambiguity when they are projected to a plane. – Alex Provost Jan 10 '16 at 21:30
  • Is there a name for the above function $\tau$ in cohomology theory? It seems to me something like curvature. Has it really anything to do with curvature? – mma Sep 05 '17 at 05:46
  • 2
    @mma Interesting question. The way Penrose used $\tau$ is to identify a 1-cocycle (the "distance cocycle") which is not a coboundary, the key fact being $\tau : C^1(X) \to \Bbb R$ being 1 on coboundaries. I suppose it's kind of like integration to me, in the same way that integral over exact forms are $0$ can be used to come up with closed forms which are not exact. – Balarka Sen Sep 05 '17 at 07:13