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From a recent answer by Mike Shulman, I read:

"HoTT is (among other things) a foundational theory, on roughly the same ontological level as ZFC, whose basic objects can be regarded as $\infty$-groupoids"

Now, $\infty$-groupoids are the same thing as spaces, and I have a couple of spaces which I like.
Here's one which I like a lot: the Lie group $SU(n)$.

Question:
How do I say "let $X:=SU(n)$" in HoTT?

Here, $n$ can be either a variable, or a definite number (say $3$).

Disclaimer: I (unfortunately) don't know much about HoTT. So please be gentle.

YCor
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    Well, for starters, $SU(2)$ is a $3$-sphere $S^3$, and my understanding is that spheres can be presented in HoTT as higher inductive types: the $3$-sphere in particular I think can be presented in a way that amounts to saying "the free $\infty$-groupoid on a $3$-automorphism" (although I don't actually know enough about HoTT to verify this). – Qiaochu Yuan Jan 02 '18 at 00:47
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    Thank you Qiaochu. Yes, I was aware of that. But $SU(2)$ is quite special, and I was conscious not to say "(say $n=2$)" in my formulation of the question. – André Henriques Jan 02 '18 at 01:27
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    Another option is to try defining $BSU(n)$ instead of $SU(n)$, then taking loop spaces. Naively (probably I know even less about HoTT than you) this comes down to defining its functor of points, so defining complex vector bundles equipped with a complex orientation. – Qiaochu Yuan Jan 02 '18 at 02:16
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    There are two big problems with defining geometric objects like $SU(n)$ in HoTT. The first is that HoTT works only up to homotopy equivalence, so many basic objects like $\mathbb R^n$ are not natively definable (other than by a verbatim reproduction of set-theoretic definitions). The second is that HoTT is the language of higher topoi, so if you can define things like K-theory, bordisms and chromatic filtration, then you know how to do it in arbitrary topos, including examples like equivariant and motivic topoi. I don't know what should be complex bordism in those categories. – Anton Fetisov Jan 02 '18 at 04:26
  • @QiaochuYuan An extremely naïve approach to $BU(n)$ would be to try to give sense to the "multiplicative quotient of $BU(1)^n$ by the $n$-permutations". I mean by that building such $BU(1)^n\to Q$ which would induce $\hom(Q,R)\to\hom(BU(1)^n,R)$ identifying the image with $\hom(BU(1)^n,R)^{\Sigma_n}$ for any (nice enough?) ring $R$. Except that probably this only makes sense in the stable context... – მამუკა ჯიბლაძე Jan 02 '18 at 05:22
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    @AntonFetisov The $\infty$-groupoid "$\mathbb{R}^n$" is contractible, so of course it's definable in HoTT. Whereas the topological space "$\mathbb{R}^n$" is definable using a "verbatim reproduction of set-theoretic definitions"; I don't know why that would be regarded as a problem. The real problem is that the relationship between the two is hard to define. – Mike Shulman Jan 02 '18 at 07:18
  • And I don't think "if you can do it in HoTT you can do it in an arbitrary topos" is a problem with defining things in HoTT; rather it's a reason to define things in HoTT, because if you succeed then you'll know the answer in any topos! Though actually, usually the answer to this sort of question just ends up lying in the constant objects of a topos, since it is constructed out of finite limits and arbitrary colimits, hence is preserved by inverse images of geometric morphisms. – Mike Shulman Jan 02 '18 at 07:20
  • @MikeShulman The fact that HoTT proofs work in any topos is both a blessing and a curse: yes, once we write a proof we get a result automatically in any context, but we are also forced to prove it in all topoi at once, which implies a much higher standard of proof than usual. In particular, it's hard to use topos-specific properties and we certainly can't rely on model-dependent statements, even if they could be useful. – Anton Fetisov Jan 02 '18 at 12:45
  • "$\mathbb R^n$ is contractible so it's definable" - this is a non-answer. The reason to define $\mathbb R^n$ is to work with structures defined over it: vector spaces, Banach algebras, moduli spaces of points, linear groups etc. None of this is possible if we blindly set $\mathbb R^n = 1$, and the roundabout way through set-theoretic definitions makes one question the use of homotopy foundations in the first place, since we'd go back to dealing with the same problems. This problem is related to the compactification of $Spec, Z$: homotopy theory is its open part, and we need fiber at $\infty$. – Anton Fetisov Jan 02 '18 at 12:48
  • @AntonFetisov Re: topoi, what I meant is that it's wrong to conclude from the generality of HoTT that "we don't know how to do X in HoTT because we don't yet know the answer to X in arbitrary topoi". That makes it sound as if a proof in HoTT were a collection of unrelated proofs, one for each topos, which is not the case. Instead, a proof in HoTT is an argument in a formal system, which by virtue of its "constructive" nature happens to be interpretable in any topos. (cont.) – Mike Shulman Jan 02 '18 at 17:41
  • So it could happen that even though we don't think we know the answer in an arbitrary topos yet, one of the "classical" proofs happens to (essentially) already be "constructive" enough to be writable in HoTT, and hence automatically give us the answer in an arbitrary topos. It's true that the generality of HoTT does make some classical arguments invalid, but when that happens it's due to intrinsic properties of the arguments themselves (use of nonconstructive principles like the axiom of choice). – Mike Shulman Jan 02 '18 at 17:43
  • And of course one wouldn't actually define $\mathbb{R}^n=1$. I mentioned that only to emphasize why it's confused to expect a "homotopical" definition of $\mathbb{R}^n$ to look any different from the set-theoretic one. If what you're working with is vector spaces and Banach algebras, then you're simply in the world of sets (0-groupoids), so there's no reason to expect your mathematics to look any different in a "homotopical foundation". HoTT is not a magic wand that "solves all your problems"; the majority of mathematics is (so far) non-homotopical and is no easier in HoTT than in ZFC. – Mike Shulman Jan 02 '18 at 17:46
  • @Mike If I'm working with vector spaces then I most certainly want them to natively carry groupoid information, since I want arbitrary vector bundles, not discrete (=flat) ones. I want the full $\infty$-category of real/complex vector spaces, I want K-theory and bordisms. The same with Banach algebras. One can certainly take the PoV that HoTT is just a nice tool for formalization and the rest of mathematics should be done as before, but I firmly believe that there is much more to be gained with proper definitions. – Anton Fetisov Jan 02 '18 at 17:54
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    That's why you should care about all topoi: a definition that doesn't advance our understanding of topoi is just a dirty hack. Of course, my PoV is that all good mathematics is based on homotopy theory, the rest either didn't catch up or isn't important. An extreme view, certainly. – Anton Fetisov Jan 02 '18 at 17:56
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    @AntonFetisov The type $\mathrm{Vect}$ of vector spaces, defined naively in HoTT, does automatically carry groupoid information, by the univalence axiom. For instance, if $S^1$ is the homotopical HIT circle, then a map $S^1 \to \mathrm{Vect}$ corresponds to a not-necessarily-flat vector bundle over the circle. If you want to relate this to the point-set topological structure on the total space of such a bundle over the topological circle, you run into the problem of fundamental $\infty$-groupoids discussed in the answers below, and which cohesive HoTT solves. – Mike Shulman Jan 02 '18 at 18:26
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    In particular, if $\mathbb{S}^1= { (x,y) \mid x^2+y^2=1}$ denotes the topological circle in cohesive HoTT, and $\mathrm{Vect}$ is the type of vector spaces defined naively, with no explicit attention paid to "topology" or "groupoid" information in either case, then it happens automatically in cohesive HoTT that maps $\mathbb{S}^1 \to \mathrm{Vect}$ represent (not necessarily flat or trivial) continuous vector bundles over the topological circle. So I believe that (cohesive) HoTT really does do what you want. – Mike Shulman Jan 02 '18 at 18:28

3 Answers3

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I think Noah's answer is mostly right, but partly misleading, and explaining why will take too much space for a comment, so I'm posting a separate answer.

As Noah says, the main conceptual point is that HoTT forces us to un-confuse ourselves about the difference between a topological space and an $\infty$-groupoid, which are conceptually distinct but historically have been conflated in algebraic topology. (Actually, higher topos theory already pushes us in that direction, but HoTT is more insistent.)

On the other hand, the main technical point is that in "Book HoTT" we don't know how to define the "fundamental $\infty$-groupoid" functor that takes a topological space to its associated $\infty$-groupoid. Two important things to emphasize about this sentence are (1) it's a statement about a particular formal system (the one used in the HoTT Book), which is just one piece of the subject area of HoTT, and (2) it's (currently) a statement about ignorance rather than impossibility. However, it does have two important upshots:

  1. In Book HoTT, to define a particular $\infty$-groupoid, you have to define it as an $\infty$-groupoid, i.e. using purely $\infty$-groupoidal notions; you can't "cheat" by defining it to be the fundamental $\infty$-groupoid of some topological space. (The definition also has to be "finitary" in some sense — it can use infinitary constructions, but they have to be finitely expressible such as by induction over the internal natural numbers.)

  2. In Book HoTT, you are out of luck if you want to use topological spaces to prove things about their fundamental $\infty$-groupoids, or use fundamental $\infty$-groupoids to prove things about their originating topological spaces.

These problems, though significant, are part of the reason I said in the linked-to answer that "the promise of HoTT is not yet fully realized". Keep in mind that HoTT is only a few years old; we are already working on various ways to solve them.

The first of them is much less of an an insurmountable obstacle than you might think. Quite often, it turns out that an $\infty$-groupoid that's classically defined from some topological space also has a very natural presentation in purely $\infty$-categorical language, which can sometimes yield important new insight about it and enables new clean proofs in synthetic homotopy theory. The simplest example is the spheres: as Qiaochu mentioned in a comment, $S^n$ can be defined as freely generated by a point and an $n$-automorphism. Any CW-complex decomposition of a space is also a free presentation of its fundamental $\infty$-groupoid, though to be expressible in HoTT such a decomposition must be finite or "finitarily definable" (by internal induction). And classifying spaces can easily be defined using the univalent universe; see this blog post for instance. And recently, Ulrik Buchholtz and Egbert Rijke have made an important advance in unifying the latter two approaches to classifying spaces, using classification information as a way to finitarily present a CW-complex decomposition; in this paper they apply it to real projective spaces, but I think similar ideas should work rather more generally.

I don't know whether anyone has written down a definition of the $\infty$-groupoid $SU(n)$ in particular, but I'm confident that it will be possible. In fact it will probably be better to directly construct its classifying space $BSU(n)$ and then define $SU(n) = \Omega BSU(n)$, since that way the group structure of $SU(n)$ will arise automatically (delooping arbitrary $\infty$-groupoids is another thing we don't know how to do inside Book HoTT). I disagree with Noah that this would "not be constructing $SU(n)$ but something else equivalent to it": it would be a perfectly valid construction of the $\infty$-groupoid $SU(n)$, and an $\infty$-groupoid is only defined up to equivalence. What it wouldn't be is a proof that that $\infty$-groupoid is the fundamental $\infty$-groupoid of some topological space. Of course the name "$SU(n)$" for this $\infty$-groupoid is derived from the topological space of which classically it is the fundamental $\infty$-groupoid, but that accident of history is no reason to denigrate the existence of the $\infty$-groupoid in its own right.

The second problem above is rather more problematic, and there are various ways one might try to solve it. To start with, one might hope to prove that fundamental $\infty$-groupoids are definable in Book HoTT; but I don't know of anyone working on this, and I think the general feeling seems to be that it's probably not possible.

The other possibility is to introduce formal systems that are better than Book HoTT. Since the obstacle to defining fundamental $\infty$-groupoids in Book HoTT is their infinitary nature (you have to say "continuous maps $D^k \to X$ induce $k$-dimensional paths in $\Pi_\infty(X)$" for all $k$, coherently), the "obvious" way to proceed is to enhance it with ways to talk about infinitary things more cleanly. Here I think the state of the art is two-level type theory, a constellation of formal systems and ideas revolving around the idea of re-introducing (in a controlled way) the sort of "strict point-set-level equality" that Book HoTT mostly eschews. I don't think anyone has just yet applied this to defining fundamental $\infty$-groupoids, but it should be quite possible.

Another possibility is cohesive HoTT, which is less obvious and requires more re-training of our intuition, but in my opinion yields a more elegant result. Contrary to the former approach (and what Noah implied), cohesive HoTT does not require you to explicitly define manifolds, topological spaces, or even fundamental $\infty$-groupoids! In fact this is precisely the point: just as "plain" HoTT is a foundational theory whose basic objects can be regarded as $\infty$-groupoids, cohesive HoTT is a foundational theory whose basic objects can be regarded as topological $\infty$-groupoids (the motivating model consists of $\infty$-stacks on the site of Euclidean spaces $\mathbb{R}^n$). So just as in plain HoTT you don't need to define all the "$k$-simplices" of $S^n$ separately but they arise automatically from its presentation via generators and relations, in cohesive HoTT you don't need to define the topology of $\mathbb{R}^n$ explicitly but it arises automatically from the definition of the set $\mathbb{R}$ using Dedekind cuts. Moreover, you don't need to define the fundamental groupoid functor by explicitly specifying its $k$-paths either; it has a simple universal property relating this sort of "intrinsic topology" to the "intrinsic $\infty$-groupoidness" of plain HoTT.

So in conclusion, in cohesive HoTT you can define the $\infty$-groupoid $SU(n)$ more easily than in classical algebraic topology. Define $\mathbb{R}$ using Dedekind cuts, define $\mathbb{C}$ in the obvious way, and define the set $SU(n) \subseteq \mathbb{C}^{n^2}$ using the usual algebraic formulas, then apply the fundamental $\infty$-groupoid functor (which in cohesive HoTT is called its "shape", written ʃ $SU(n)$). You don't need to think about the topology of $\mathbb{C}$ or the manifold structure of $SU(n)$ at all; that gets carried along for the ride automatically. (You do, however, need to be a little careful to do things in the correct "constructive" way, which for instance means using the correct constructive notion of "Dedekind cut".)

Mike Shulman
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    Thank you Mike for this very informative answer. Really, what I wanted to know is how one defines $SU(n)$. As far as I can tell, you've addressed that question is the last paragraph only. But it went a bit quickly. I would appreciate if you could expand (maybe in a different answer) what is meant by (1) "Define ℝ using Dedekind cuts", (2) "define ℂ in the obvious way" (3) "define the set SU(n) using the usual algebraic formulas", and (4) "apply the fundamental ∞-groupoid functor". In particular, does step (3) require n to be a definite number, or can I take it to be a variable? – André Henriques Jan 02 '18 at 13:21
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    Another request for clarification: From your words "I don't know whether anyone has written down a definition of the ∞-groupoid SU(n) in particular, but I'm confident that it will be possible", I'm not able to tell whether [a] this is something which is in principle easy, but no one happens to have done it, or [b] this is something for which you have essentially no idea how to do it, but you're optimistic about the future. Could you please comment. – André Henriques Jan 02 '18 at 13:24
  • Thanks! This is very clarifying. I should have been more clear about the downside of giving an “algebraic construction”, what I meant was that if you just find by some long calculation an opaque CW description of the $\infty$-groupoid of $SU(n)$ that probably won’t be useful in that you can’t easily do with it anything you want to do with it. If there’s a construction directly as an $\infty$-groupoid which captures the main uses (group structure, principal bundles, fibrations with SU(n-1), etc.) then yes that’s great. – Noah Snyder Jan 02 '18 at 14:34
  • Treating n as a variable should be easy. Step 1 is in the HoTT book (and recalled in Mike’s paper). Step 4 is a big part of the content of Mike’s paper that I linked to. Steps 2 and 3 are straightforward (they’re literally the usual definitions). Don’t know about your other question – Noah Snyder Jan 02 '18 at 15:12
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    @AndréHenriques Noah's answer to the question about the cohesive approach is correct. For instance, $\mathbb{C} = \mathbb{R}\times\mathbb{R}$ with $(x,y)\cdot (u,v) = (x u - y v , x v + y u)$ and so on. Since this is all happening at set-level (before step (4)), there is no coherence to worry about, and one can treat $n$ as a variable in type theory just the way one does it in set theory. – Mike Shulman Jan 02 '18 at 17:00
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    As for the $\infty$-groupoid $SU(n)$, I would say the situtaion is in between [a] and [b]. Before the Buchholtz-Rijke paper I would have said [b], but now we have a general technique that should be applicable to constructions of this sort, so I would say it's no longer the case that we have "no idea" how to do it: we do have ideas. However, neither would I say it's "easy" since some additional thought and ideas will probably be required: it'll be a new application of a not-yet-fully-explored technique. (It could also be that I'm wrong and their technique won't help so much.) – Mike Shulman Jan 02 '18 at 17:02
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    There's a short description of our HoTT version of the Milnor construction of projective spaces for any $\infty$-group in these talk notes. We thought a bit about using the Schubert calculus to define at least the types SU(n), but one really wants the deloopings. It's still an open problem to define the delooping BSU(2), or any other delooping of $S^3$, in Book HoTT. Egbert has a different approach using $\infty$-equivalence relations, but no construction yet. I like Mike's cohesive approach as the most elegant way. – Ulrik Buchholtz Jan 02 '18 at 17:22
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    By the way, I regret that my real-cohesion paper is so long. It's that long because I tried to be fairly comprehensive about describing the behavior of the modalities, so the paper could function as a reference for future work in cohesive HoTT. But the basic ideas can be described much more briefly; check out the long and chatty introduction to the paper, and these talk slides: http://home.sandiego.edu/~shulman/papers/realcohesion-cmu2017.pdf . – Mike Shulman Jan 02 '18 at 18:22
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This isn't easy to do, and the reason it isn't easy is because of the step "$\infty$-groupoids are the same thing as spaces." Of course the homotopy hypothesis tells you that any $\infty$-groupoid is equivalent to the fundamental $\infty$-groupoid of a space, but that doesn't mean that they're literally the same thing. The way that HoTT approaches about $\infty$-groupoids is not by thinking of them as spaces. What's easy in HoTT is to do something like, "consider the $\infty$-groupoid freely generated by a single point and a single 1-path from that point to itself." This $\infty$-groupoid is equivalent to the fundamental $\infty$-groupoid of classical circle, but we haven't constructed it that way, instead we've constructed it algebraically via generators/relations.

Ok, so the first thing you could do is find an "algebraic" description of the fundamental $\infty$-groupoid of $\mathrm{SU}(n)$. Essentially this is giving an explicit CW-decomposition of $\mathrm{SU}(n)$. I'm not sure how hard this is off the top of my head, but one needs to be a bit careful with such an approach to make sure to get something that's just as useful as $\mathrm{SU}(n)$ in practice. For example, as suggested in comments it would be better to construct $B\mathrm{SU}(n)$ instead so that when you take the loop space you know you have a group structure. But you also want to be able to handle things like the relationship between $\mathrm{SU}(n)$ and complex vector bundles, and if your construction is sufficiently messy and computational then it might not be good enough.

Instead, you could try to build up the theory of manifolds (or something similar) inside HoTT and then define a notion of fundamental $\infty$-groupoid. If I understand things right, this is something that real cohesive type theory will let you do. But there's a lot you need to do in order to get there, for a first step you need to define $\mathbb{R}$ inside HoTT which has some tricky points because HoTT is naturally constructive, but the Cauchy reals and the Dedekind reals are different constructively. In HoTT $\infty$-groupoids are foundational native objects, but $\mathbb{R}$ is still just as complicated (well, slightly more complicated) as it was classically. From there you can start trying to define manifolds (or some more general notion of a topological space of similar flavor to manifolds) and then defining the fundamental $\infty$-groupoid of such a space. All of this is done in Mike's paper where he develops a type-theory for topological $\infty$-groupoids. The theory there is developed to the point where he shows that $\{(x,y)\in \mathbb{R}^2: x^2+y^2=1\}$ is something that you can take the fundamental $\infty$-groupoid (or "shape") which outputs an $\infty$-groupoid and that the answer you get is equivalent to the type-theoretic circle. If I understand everything correctly, you should be able to use this to define the shape of any real algebraic manifold (including $\mathrm{SU}(n)$), as well as many other "geometrically" defined topological spaces.

In conclusion, HoTT does let you deal with $\infty$-groupoids nicely and directly, but doesn't easily give you access to $\infty$-groupoids that are of a topological nature rather than an algebraic nature. In order to access those you still need to build up real analysis and topology or some replacement for them, which takes a bit of work (as it does classically) and also has a few tricky points due to the constructive nature of HoTT.

Noah Snyder
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    If this is true, then this sounds like a HUGE drawback of HoTT. One of the strengths of algebraic topology is that spaces come from so many different types of constructions: there's Eilenberg MacLane space (among other spaces defined via their Postnikov towers), there's spheres (and other spaces defined via their cell structures), but then there's many more spaces which are defined by ways which are not directly related to their Postnikov towers, nor to their cell structures. $SU(n)$ is just the first example: I would call them "God given spaces" – André Henriques Jan 02 '18 at 01:44
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    I'd like to point and that it's by relying on the information given by those "God given spaces" that some of the deepest results of homotopy theory have been achieved. For example, the chromatic filtration in stable homotopy theory relies heavily on the existence of complex cobordism $MU$. And the latter is constructed by using, at its very base, the unitary groups... Similarly for $K$-theory: it's constructed using the unitary groups. So if we can't define the groups $U(n)$, then that sounds really bad. – André Henriques Jan 02 '18 at 01:48
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    Yes it's a drawback, and one certainly needs to eventually be able to talk about $SU(n)$. No approach is going to make everything easy, and the price of making some things easier is making other things harder. So I don't think it's terrible that talking about $SU(n)$ is harder in HoTT than it is from a topological viewpoint, so long as it's not too hard. – Noah Snyder Jan 02 '18 at 01:52
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    I really shouldn't rant before learning the subject (I use this opportunity to apologise). The answer to my question might actually be contained in what you wrote, in the form of real cohesive type theory. So my question then becomes: "How does one define $SU(n)$ in real cohesive type theory?" Or possibly: "Is real cohesive type theory already developed enough so that one can use it to define $SU(n)$?" – André Henriques Jan 02 '18 at 01:57
  • Deleted a few of my comments responding to Andre's last comment and incorporated them into the text of the answer. – Noah Snyder Jan 02 '18 at 02:42
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    This answer reinforces my outsider's impression that HoTT is more oriented toward making the theory easier than illuminating concrete examples - I regret lacking the courage to ask a naive question like this one. But you also make it sound as though there are other examples which are more natural in this language; do these examples correspond to familiar constructions in "ordinary" mathematics? – Paul Siegel Jan 02 '18 at 07:52
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    My comment is much more than 600 characters, so I wrote it as a separate answer. – Mike Shulman Jan 02 '18 at 08:12
  • I'd have accepted the other answer. Since this is now the top answer I'm going to edit the second paragraph to make it clearer without reading the comments on Mike's answer. – Noah Snyder Jan 03 '18 at 16:47
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As time passes I've become increasingly unhappy with my original answer, so I want to give a new one.

One of the most interesting things about HoTT is that it lets you ask questions of a logical flavor about homotopy theory. For example, I was on Robert Rose's defense committee where the thesis problem was suppose you look at a version of HoTT where you don't assume you have the natural numbers, but you do have the circle, what can you still say? Can you still show that you have infinite sets? In what sense of the word infinite? How much number theory do you still get? These are interesting questions about the logic of homotopy theory that you can't even ask in the usual space-based foundations (classically how are you going to even define the circle without having access to infinite sets?).

So what I'd say now is that the question of whether (and how) you can define BSU(n) in book HoTT is an extremely interesting "logical" question about homotopy theory. As Mike explains you can certainly do this if you give yourself a stronger type theory than book HoTT, so if your goal is to just give yourself these tools you can do so, but it's still interesting to ask whether you can do this with the weaker book HoTT.

In particular, as you well know, classically there's a homotopy coherent action of SO(n) on n-loops via the geometric action on $E_n$-algebras. But this definition is highly geometric, which makes it difficult in practice to understand what this action actually looks like. For example, you might want to understand the action of $SO(4)$ on 4-loops in an $\infty$-groupoid well enough to be able to write down the generator of $\pi_7(S^4)$ algebraically. Furthermore, you might want to know whether you can keep doing this in general. The question of whether you can define the group $SO(n)$ in book HoTT and define its action on $\Omega^n S^n$ in book HoTT gives you a logical framework for asking this question! So the fact that you can't easily define the group $SO(n)$ in book HoTT is a feature, not a bug, because it lets you precisely phrase this very interesting mathematical question. (Of course, since it's an interesting difficult question, simply stating the question doesn't get you very far towards solving it.)

Noah Snyder
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