Equality vs. isomorphism vs. specific isomorphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:21:42Z http://mathoverflow.net/feeds/question/24378 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism Equality vs. isomorphism vs. specific isomorphism Minhyong Kim 2010-05-12T12:02:18Z 2010-05-14T10:00:41Z <p><a href="http://mathoverflow.net/questions/24318/are-there-situations-when-regarding-isomorphic-objects-as-identical-leads-to-mist" rel="nofollow">This question</a> prompted a reformulation: </p> <p><em>What is a really good example of a situation where keeping track of isomorphisms leads to tangible benefit?</em></p> <p>I believe this to be a serious question because it actually is oftentimes a good idea casually to identify isomorphism classes. To bring up an intermediate-level example I've alluded to often, consider the classification of topological surfaces. When I explain it to students, I do somewhat consciously write equalities as I manipulate one shape into another homeomorphic one. I even do it rather quickly to encourage intuitive associations that are likely to be useful. In any case, for arguments of that sort, it would be really tedious, and probably pointless, to write down isomorphisms with any precision.</p> <p>Meanwhile, at other times, I've also joined in the chorus of criticism that greets the conflation of equality and isomorphism.</p> <p>The problem is it's quite challenging to come up with really striking examples where this care is rewarded. Let me start off with a somewhat specialized class of examples. These come from <em>descent theory</em>. The setting is a map $$X\rightarrow Y,$$ which is usually submersive, in some sense suitable to the situation. You would like criteria for an object $V$ lying over $X$, say a fiber bundle, to arise as a pull-back of an object on $Y$. There is a range of formalism to deal with this problem, but I'll just mention two cases. One is when $Y=X/G$, the orbit space of a group action on $X$. For $V$ to be pulled-back from $Y$, we should have $g^*(V)\simeq V$ for each $g\in G$. But that's not enough. What is actually required is that there be a collection of isomorphisms $$f_g: g^*(V)\simeq V$$ that are compatible with the group structure. This means something like $$f_{gh}=f_g\circ f_h,$$ except you have to twist in an obvious way to take into account the correct domain. So you see, I have at least to introduce notation for the isomorphisms involved to formulate the right condition. In practice, when you want to <em>construct</em> something on $Y$ starting from something on $X$, you have to specify the $f_g$ rather precisely. </p> <p>Another elementary case is when $X$ is an open covering $(U_i)$ of $Y$. Then an object on $Y$ is typically equivalent to a collection $V_i$ of objects, one on each $U_i$, but with additional data. Here as well, $V_i$ and $V_j$ obviously have to agree on the intersections. But that's again not enough. Rather there should be a collection of isomorphisms $$\phi_{ji}: V_i|U_i\cap U_j\simeq V_j|U_i\cap U_j$$ that are compatible on the triple overlaps: $$\phi_{kj}\circ \phi_{ji}=\phi_{ki}.$$ Incidentally, for something like a vector bundle, since any two of the same rank are locally 'the same,' it's clear that keeping track of isomorphisms will be the key to the transition from collections of local objects to a global object. The formalism is concretely applied in situations where you can define some objects only locally, but would like to glue them together to get a global object. For a really definite example that comes immediately to mind, there is the determinant of cohomology for vector bundles on a family of varieties over a parameter space $Y$. Because a choice of resolution is involved in defining this determinant, which might exist only locally on $Y$, Knudsen and Mumford struggled quite a bit to show that the local constructions glue together. Then Grothendieck suggested the remedy of defining the determinant provisionally as a <em>signed</em> line bundle, which then allowed them to nail down the correct $\phi_{ji}$. These days, this determinant is a very widely useful tool, for example, in generating line bundles on moduli spaces.</p> <p>I apologize if this last paragraph is a bit too convoluted for non-specialists. Part of my reason for writing it down is to illustrate that my main examples for bolstering the 'keep track of isomorphisms' paradigm are a bit too advanced for most undergraduates.</p> <p>So, to conclude, I'd be quite happy to hear of better examples. As already suggested above, it would be nice to have them be accessible but substantively illuminating. If you would like to discuss, say, different bases for vector spaces, it would be good if the language of isomorphism etc. clarifies matters in a really obvious way, as opposed to a sets-and-elements exposition.</p> <p>Added: Oh, if you have advanced examples, I would certainly like to hear about them as well.</p> <p>Added: I see now there are three levels at least to distinguish:</p> <p>Regarding objects as equal vs. regarding them as isomorphic vs. paying attention to specific isomorphisms.</p> <p>I somehow conflated the two transitions in the course of asking the question. Of course I'm happy to see good examples illustrating the nature of either, but I'm especially interested in the second refinement.</p> <p>Added yet again:: I'm grateful to everyone for contributing nice examples, and to Urs Schreiber who put in some effort to instruct me over at the <a href="http://golem.ph.utexas.edu/category/2009/10/math_overflow.html#comments" rel="nofollow">n-category cafe</a>. As I mentioned to Urs there, it would be especially nice to see examples of the following sort.</p> <ol> <li><p>One usually thinks $X=Y$;</p></li> <li><p>A careful analysis encourages the view $X\simeq Y$;</p></li> <li><p>This perspective leads to genuinely new insight and benefit. </p></li> </ol> <p>Even better would be if some specific knowledge of the isomorphism in 2. is important. Of course, more than two objects might be involved. I was initially hoping for some input from combinatorics, with the emphasis on 'bijective proofs' and all that. Anything?</p> <p>Added, 14 May:</p> <p>OK, I hope this will be the last addition. Because this question flowed over to the n-category cafe, I ended up having a small discussion there as well. I thought I'd copy here my last response, in case anyone else is interested.</p> <p>n-cafe post:</p> <p>I suppose it's obvious by now that I'm using a specific request to drive home the need for 'small but striking examples' in favor of category theory.</p> <p>Last fall, Eugenia Cheng told me of a visit to some university to give a colloquium talk. The host greeted her with the observation that he doesn't regard category theory as a field of research. OK, he was probably a bit extreme, but milder versions of that view are quite common. Now, one possible response is to regard all such people as unreasonable and talk just to friends (who of course are the reasonable people!). This is not entirely bad, because that might be a way to buy time and gain enough stability to eventually prove the earth-shattering result that will show everyone! Another way is to take up the skepticism as a constructive everyday challenge. This I suppose is what everyone here is doing at some level, anyways.</p> <p>Other than the derived loop space, which is not exactly small, Urs' examples are all of the simple subtle sort that can, over time, contribute to a really important change in scientific outlook and maybe even the infrastructure of a truly glorious theory. For example, I agree wholeheartedly about the horrors of the old tensor formalism. But it's not unreasonable to ask for more striking accessible evidence of utility when it comes to the current state of category theory.</p> <p>The importance of small insights and language that gradually accumulate into the edifice of a coherent and powerful theory is the usual interpretation of Grothendieck's 'rising sea' philosophy. However, the process is hardly ever smooth along the way, especially the question of acceptance by the community. I'm not a historian, but I've studied arithmetic geometry long enough to have some sense of the changing climate surrounding etale cohomology theory, for example, over the last several decades. The full proof of the Weil conjectures took a while to come about, as you know. Acceptance came slowly with many bits and pieces sporadically giving people the sense that all those subtleties and abstractions are really worthwhile. Fortunately, the rationality of the zeta function was proved early on. However, there was a pretty concrete earlier proof of that as well using $p$-adic analysis, so I doubt it would have been the big theorem that convinced everyone. One real breakthrough came in the late sixties when Deligne used etale cohomology to show that Ramanujan's conjecture on his tau function could be reduced to the Weil conjectures. There was no way to do this without etale cohomology and the conjecture in question concerned something very precise, the growth rate of natural arithmetic functions. This could even be checked numerically, so impressed people in the same way that experimental verification of a theoretical prediction does in physics. Clearly something deep was going on. Of course there were many other indications. The construction of entirely new representations of the Galois group of $\mathbb{Q}$ with very rich properties, the unification of Galois cohomology and topological cohomology, a clean interpretation of arithmetic duality theorems that gave a re-interpretation of class field theory, and so on.</p> <p>For myself, being a fan of you folks here, I believe this kind of process is going on in category theory. But I don't think you have to be too unreasonable to doubt it. In a similar vein, I don't agree with Andrew Wiles' view that physics will be irrelevant for number theory, but also think his pessimism is perfectly sensible.</p> <p>I think I'm trying to make the obvious point that the presence of pessimists can be very helpful to the development of a theory, in so far as the optimists interact with them in constructive ways. I haven't been coming to this site much lately, because the bit of internet time I have tends to be absorbed by Math Overflow. But I did catch David's <a href="http://golem.ph.utexas.edu/category/2010/05/quinn_on_higherdimensional_alg.html#more" rel="nofollow">recent post</a> on Frank Quinn's article, which ended up as a catalyst for my MO question.</p> <p>At the Boston conference following the proof of Fermat's last theorem, I've been told Hendrik Lenstra said something like this: 'When I was young, I knew I wanted to solve Diophantine equations. I also knew I didn't want to represent functors. Now I have to represent functors to solve Diophantine equations!' So should we conclude that he was foolish to avoid representable functors for so long? I wouldn't.</p> <p><a href="http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24563#24563" rel="nofollow">This response</a> to the MO question brings up the importance of knowing the specific isomorphism between some Hilbert spaces given by the Fourier transform. This is an excellent example, especially when we consider how it relates to the different realizations of the representations of the Heisenberg group and the attendant global issues, say as you vary over a family of polarizations. But I couldn't resist recalling Irving Segal's insistence that 'There's only <em>one</em> Hilbert space!' Obviously, he knew, among many other things, the different realizations of the Stone-Von-Neumann representation as well as anyone, so you can take your own guess as to the reasoning behind that proclamation. He certainly may have lost something through that kind of philosophical intransigence. But I suspect that he, and many around him, gained something as well.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24381#24381 Answer by Joel David Hamkins for Equality vs. isomorphism vs. specific isomorphism Joel David Hamkins 2010-05-12T12:23:01Z 2010-05-12T12:23:01Z <p>Automorphism groups are studied intensively in mathematics, and these groups explicitly track the difference between isomorphisms and equality. We aren't willing to say that every automorphism of a mathematical structure is really just the equality map, since under such a perspective the automorphism group would evaporate.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24383#24383 Answer by Neel Krishnaswami for Equality vs. isomorphism vs. specific isomorphism Neel Krishnaswami 2010-05-12T12:56:31Z 2010-05-12T12:56:31Z <p>Natural examples arise in proof theory: if you equate isomorphic propositions, then the semantics of proofs collapses -- all proofs of a given proposition end up having the same denotation. </p> <p>But if you avoid identifying isomorphic propositions, then lattices naturally flower into categories, with the morphisms embodying the semantics of proofs. (If you like, you can view model theory as performing the collapse on purpose, and then studying what is left over.)</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24387#24387 Answer by Chandan Singh Dalawat for Equality vs. isomorphism vs. specific isomorphism Chandan Singh Dalawat 2010-05-12T13:10:34Z 2010-05-12T13:10:34Z <p>Hi Minhyong,</p> <p>I don't know whether this example throws some light on your question, but here it is :</p> <p>Let $K$ be a finite extension of ${\bf Q}_p$, and suppose you want to count the number of degree-$p$ extensions of $K$, of which there are only finitely many, as you know.</p> <p>But counting them up to $K$-isomorphism is one thing, and counting the actual number of extensions in a given algebraic closure is quite another. In his mass formula, Serre did the latter count, with some weights coming from the ramification : when $E$ runs through ramified degree-$p$ extensions of $K$,<br> $$\sum_E q^{-c(E)} =p,$$ where $c(E)=v(d_{E|K})-(p-1)$, $v(d_{E|K})>(p-1)$ is the valuation of the discriminant $d_{E|K}$ of $E|K$, and $q$ is the residual cardinality of $K$.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24395#24395 Answer by Simon Thomas for Equality vs. isomorphism vs. specific isomorphism Simon Thomas 2010-05-12T14:27:32Z 2010-05-12T14:45:58Z <p>Perhaps another reason to distinguish between isomorphic structures and isomorphism classes ... Many years ago, Grigorchuk defined a very beautiful space $\mathcal{G}$ of finitely generated groups with the following features:</p> <p>(i) $\mathcal{G}$ is a Polish space; i.e. a separable completely metrizable space.</p> <p>(ii) The isomorphism relation on $\mathcal{G}$ is induced by a natural action of a suitable countable group $G$.</p> <p>However, Champetier has shown that the "quotient space" $\mathcal{G}/G$ of isomorphism types is horribly complex and cannot be given the structure of a decent topological space in any natural manner. </p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24398#24398 Answer by André Henriques for Equality vs. isomorphism vs. specific isomorphism André Henriques 2010-05-12T14:34:28Z 2010-05-12T16:20:15Z <p>An example:</p> <p>In homotopy theory, it is useful to consider the homotopy groups of spaces. These are <em>isomorphism classes</em> of maps $S^n\to X$.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24405#24405 Answer by Mike Shulman for Equality vs. isomorphism vs. specific isomorphism Mike Shulman 2010-05-12T15:26:13Z 2010-05-12T15:26:13Z <p>I think a great example of this is <a href="http://ncatlab.org/nlab/show/groupoid%20cardinality" rel="nofollow">groupoid cardinality</a>. The cardinality of a groupoid $G$ is defined to be $$\sum_{[x] \in \pi_0(G)} \frac{1}{|Aut(x)|}$$ If the groupoid is a discrete set, this reduces to the usual notion of cardinality, while in general things with more automorphisms are counted as "less than one." This may seem weird until you start seeing it pop up all over the place. For instance, if a finite group $G$ acts on a finite set $X$, then the groupoid cardinality of the action groupoid (a.k.a. the "homotopy quotient") $X// G$ is equal to $|X|/|G|$. Groupoid cardinality also has links to exponential generating functions, via the theory of "species of structure."</p> <p>But I think the wackiest place this comes up is in <em>Feynman integrals</em>. When physicists want to evaluate a Feynman integral, they expand it out in a power series in terms of some parameter. (The power series may not converge, but that doesn't stop them from evaluating its first few terms as an approximation.) The coefficients of each term are derived from the various different ways of distributing some number of derivatives across a product of different terms, which if you work it out combinatorially can equally well be calculated by counting "Feynman diagrams" -- certain graphs which describe different ways of connecting up nodes (which it is easy to start thinking of as "pictures" of a "particle interaction"). But the kicker is that you have to count these Feynman diagrams <em>with automorphisms</em>, i.e. instead of the cardinality of the set of Feynman diagrams, the coefficient in the power series comes from the cardinality of the <em>groupoid</em> of Feynman diagrams.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24407#24407 Answer by Mike Shulman for Equality vs. isomorphism vs. specific isomorphism Mike Shulman 2010-05-12T15:37:22Z 2010-05-12T15:37:22Z <p>Suppose you have two categories C and D, and functors $F:C\to D$ and $G:D\to C$ such that for all $x\in C$, $G(F(x))$ is isomorphic to $x$, and for all $y\in D$, $F(G(y))$ is isomorphic to $y$. If you didn't know how important it was to distinguish between "isomorphic" and "an isomorphism," you might think that then C and D are essentially the same category. But of course in order for C and D to be equivalent, one additionally needs the isomorphisms in question to be <em>natural</em>.</p> <p>A nice example of a pair of functor with the above properties, but which are not an equivalence of categories, are the functors relating vector spaces and <a href="http://ncatlab.org/nlab/show/affine%20space" rel="nofollow">affine spaces</a>. Here $F:\mathrm{Vect}\to \mathrm{Aff}$ regards a vector space as an affine space by forgetting its origin, and $G:\mathrm{Aff}\to \mathrm{Vect}$ constructs the "vector space of displacements" in an affine space. The composite $G F$ is naturally isomorphic to the identity, but $F G$ is only "unnaturally" isomorphic to the identity, and the categories are definitely not equivalent. A simpler version of this example relates groups with <a href="http://ncatlab.org/nlab/show/heap" rel="nofollow">heaps</a>.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24425#24425 Answer by David Sprehn for Equality vs. isomorphism vs. specific isomorphism David Sprehn 2010-05-12T19:08:33Z 2010-05-12T19:08:33Z <p>Whitehead's Theorem states that a map between two connected CW complexes which induces isomorphisms on all homotopy groups is a homotopy equivalence. The naive reformulation which forgets to keep track of the specific isomorphisms (i.e. "Two connected CW complexes with all homotopy groups isomorphic are homotopy equivalent") is definitely false.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24439#24439 Answer by Thomas Kragh for Equality vs. isomorphism vs. specific isomorphism Thomas Kragh 2010-05-12T21:38:07Z 2010-05-12T21:38:07Z <p>How $X \times Y$ is isomorphic to $Y \times X$ in a category with products should not be ignored. To illuminate this think a little more general:</p> <p>Keeping track of isomorphisms are essential in symmetric monoidal categories:</p> <p>In a monoidal category you have a product $\oplus$ and coherency isomorphisms relating the associativity of the product $c_{A,B,C} \colon A\oplus (B \oplus C) \to (A \oplus B) \oplus C$ satisfying the pentagon relation. You also have a unit (zero) object and other isomorphisms satisfying some relations.</p> <p>There is a theorem saying that one can always replace this by an equivalent monoidal category where all these isomorphisms are identities (including the ones for the unit making the unit strict). So really this tells you that you dont need these isomorphisms to define them up to equivalence.</p> <p>However, for symmetric monoidal categories, which also have a "commutativity" isomorphism you cannot always do this, and so it is essential to keep track of the isomorphisms to ensure full generality.</p> <p>This is ture also for categorical products and coproducts, and more generally for compatibility betweem objects defined by different universal properties in a category. E.g. the eaxmple $X \times Y$ is isomorphic to $Y\times X$ mentioned at the beginning. This is more than keeping track of the automorphisms of objects (which is also very important, but mentioned in other answers) - it is remembering certain important isomorphisms depending on some objects.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24482#24482 Answer by Tom Ellis for Equality vs. isomorphism vs. specific isomorphism Tom Ellis 2010-05-13T08:56:26Z 2010-05-13T08:56:26Z <p>I have been thinking of determinants recently, so an example from multilinear algebra might help.</p> <p>Given an $n$-dimensinonal vector space $V$, the exterior products $Λ^0 V$ and $Λ^n V$ are both 1-dimensional, so are isomorphic. Moreover there is a natural isomorphism between $End(Λ^0V)$ and $End(Λ^nV)$.</p> <p>However, the maps that $A \in End(V)$ induces on $Λ^0V$ and $Λ^nV$ are different, $I$, the identity, in the former case, and $(detA)I$ in the latter.</p> <p>Perhaps what you’re looking for is not a case when keeping track of the isomorphisms matters, but when relationships with external constructions differ despite the fact that two things are isomorphic when considered in isolation.</p> <p>Previously posted here: <a href="http://golem.ph.utexas.edu/category/2009/10/math_overflow.html#c033054" rel="nofollow">http://golem.ph.utexas.edu/category/2009/10/math_overflow.html#c033054</a></p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24518#24518 Answer by Chris Schommer-Pries for Equality vs. isomorphism vs. specific isomorphism Chris Schommer-Pries 2010-05-13T16:23:27Z 2010-05-13T16:23:27Z <p>Here is an example out of algebraic topology. It is not so much about isomorphisms as about maps, but I think it is similar in spirit to what you are asking. It also might be too far beyond an undergraduate level, but I think it is a nice example where both the first and second distinctions play a role. </p> <p>The example is obstruction theory. Suppose that you have two spaces X and Y and they have maps to a third space Z. Suppose that we want to understand (homotopy classes of) lifts of the map X -> Z to a map X -> Y. Obstruction theory procedes by building a tower of spaces over Z </p> <p>$$\cdots \to Y_3 \to Y_2 \to Y_1 \to Y_0 = Z$$ such that Y is the limit of this tower, and such that the passage between layers is easier to understand. For example it might be the case that $Y_{i+1}$ is a fibration over $Y_i$ with fiber a $K(\pi, i+1)$. </p> <p>Suppose that you have lifted to a map $X \to Y_k$. The Obstruction Theory Machine then constructs for you an obstruction class $o_k$. If this vanishes then you can lift to the next level. But you usually can't define the next obstruction $o_{k+1}$ unless you remember the (homotopy class) of the lift itself. And this next obstruction depends on the choice of lift you make. If you make a bad choice, you can run into a dead end, even if other choices allow you to continue. That's a place where the first distinction plays a role. </p> <p>But now suppose that you've made choices all the way up, and all obstructions vanished along the way. Do you get a map to Y? can you classify all lifts this way? Not necessarily. There are two sets at play here (with a map between them): </p> <p>$$\pi_0 Map( X, Y) = \pi_0 Map( X, \lim Y_i ) \to \lim \pi_0Map(X, Y_i)$$</p> <p>What we have constructed is something on the right-hand-side, a coherent system of homotopy classes of lifts through the tower. What we want is the left-hand-side. We are trying to commute the functor $$\pi_0 Maps(X, -)$$ with an inverse limit, and if the tower is infinite we may not be able to do this (it is a left adjoint so commutes with colimits, not limits). If Y and all the $Y_i$ are commutative H-spaces so that these sets are abelian groups, then there will be a $\lim^1$-term. The situation is similar in the non-abelian setting. </p> <p>However if you remember not just single (homotopy classes of) maps but the whole mapping space, then this problem goes away. In other words you are saved if you remember the second level of distinction. Assuming that our spaces are reasonable we get a homotopy equivalence of mapping spaces (here my limits are homotopy limits):</p> <p>$$\underline{Map}( X, Y) \simeq \underline{Map}( X, \lim Y_i ) \to \lim \underline{Map}(X, Y_i).$$</p> <p>[All of these mapping spaces are taken relative to the map $X \to Z$, but including that in the notation would make things messy.]</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24563#24563 Answer by TS for Equality vs. isomorphism vs. specific isomorphism TS 2010-05-14T01:23:53Z 2010-05-14T01:23:53Z <p>This specific example may not count, as I don't know if anyone really views all separable infinite dimensional Hilbert spaces as being the same. But I imagine that just making the identification $L^2(0, 2\pi)=l^2(\mathbb{Z})$ based on the fact that all separable infinite Hilbert spaces are isomorphic doesn't gain you as much insight as knowing that the Fourier transform from $L^2(0, 2\pi)$ to $l^2(\mathbb{Z})$ is an isomorphism.</p>