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I've learned some differential geometry, and I notice that in many situations, people will always use the word "(canonical) identify". For example (in the following, $M$ is smooth manifold, $p\in M$:

  1. In the definition of tangent space. One way is to define the tangent space at $p$ be the vector space of derivation over $C^\infty(M)$. The other way is to define it as the equivalent class of $$ \{\gamma:(-\epsilon,\epsilon)\to M\mid \gamma \ \text{smooth and }\gamma(0)=p\}/\sim, $$ where $$ \gamma_1\sim\gamma_2:\Leftrightarrow \exists\varphi:U\to V(\subseteq \mathbb{R}^n),\text{s.t.} (\varphi\circ\gamma_1)'(0)=(\varphi\circ\gamma_2)'(0) $$ Then we say these two definitions are equivalent. I know that they are indeed isomorphic (as vector space). But in many books, authors sometimes use one definition and then another, for different purposes. Won't this cause logical confusion?
  2. The vector space and its second dual. The contravariant tensorfield of rank $1$ over $M$, i.e. a linear map $T^*M\to\mathbb{R}$, is by definition a map which sends a $1$-form to a number. But it is always “identified” with $\Gamma(TM)$, the vector field over $M$ itself. If we go back to linear algebra, this is the problem of $V^{**}\cong V$. I can understand this identification from some category point of view. But when we do this in differential geometry, do we really not need to make a distinction?
  3. If $N$ is a submanifold of $M$ with the same dimension, then $T_pS=T_pM$. But strictly speaking (if we use the definition of tangent space as derivation), the elements in $T_pS$ is derivation over $C^\infty(N)$, while $T_pM$ is over $C^\infty(M)$. Though the difference is trivial, but indeed they are not the same. Strictly, we should have $i_{*,p}T_pN=T_pM$. But the books always just simply write $T_pN=T_pM$ and do not make any distinction. Is this really relieable?
  4. If $V$ is a vector space and its own have a structure of smooth manifold, then $T_pV=V$. But why this? It's obvious that they are isomorphic because of the same dimension, but the LHS is some derivation of $C^\infty(V)$, and RHS is just $V$ itself. Why it makes sense here?

There are also many other such identifications which cause a lot of confusion in my study. Especially in a proof the other make many such identifications, then I sometimes really can't revert these proofs back to their original form without using these "identifications". Is there some suggestions or clarifications?

Gao Minghao
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    There is a very natural isomorphism from $V$ to $T_p(V)$, don't argue with dimension here. – Martin Brandenburg Sep 01 '24 at 23:54
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    When (as in all your examples) we have a specific, natural isomorphism between two things, then, when we say we "identify" these two things, we're making the convention to not write that isomorphism explicitly. We trust the reader to see where the isomorphism would be needed in our work and to mentally insert it if (s)he doen't want to use identifications. (If, on the other hand, there are isomorphisms between two things, but no particularly natural one, then we must explicitly say which isomorphism we intend to use for the identification. See also Martin Brandenburg's answer.) – Andreas Blass Sep 02 '24 at 16:59

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This happens all over the place in mathematics, not just differential geometry, and it seems that professional mathematicians get over this by assuming that a) everybody remembers or knows what these identifications are, b) these identifications are compatible with each other.

I find this very problematic since, even though most mathematicians won't admit this, a) is simply not true and puts a lot of extra unnecessary work to the readers of mathematical texts, especially students, and b) is never checked at all since, by the definition of these so-called identifications, they are never written down again, because they are supposed to be so natural, obvious etc. Your examples already show that this is far from being true. Extending a derivation from a submanifold is not obvious at all.

The real problem arises when we identify A with B and B with C, but then the identification between A and C is not the one we think of, so we will make mistakes. This happens in publications as well.

Here is a very basic example to show what I mean. In a direct product of two groups $G \times H$ we are taught that $G$ can be "identified" with a subgroup of $G \times H$. Likewise $H$. Ok but what happens when $G = H$? Now $G$ is identified with two subgroups at the same time, so are they the same? Surely not, which doesn't sound like it's a good idea to name these subgroups "identified". Ok there are two then, but which one do we take? $G \times 1$ or $1 \times G$? Or even the diagonal (which is also naturally isomorphic to $G$, after all)? Of course this specific example will not cause serious problems, I just wanted to explain in principle what might and also does go wrong.

Put differently, an "identification" should not be thought of as a relation between two objects (identified yes or no), but rather as a specific and "nice" isomorphism between objects. And there can be many isomorphisms, so we have to say which one we choose.

For all these reasons, in my work, I avoid "identifications" if possible and instead work with what they really are, isomorphisms. They are written down or referenced, and compatibilities are then verified.

This also means that the two objects which commonly are "identified" will not be identified at all. This is not possible anyway and causes confusion all the time.

In your example, we have two different(!) vector spaces attached to a point of a manifold, one vector space of curves, one vector space of derivations. Yes, there is a natural isomorphism between these, but this doesn't make them the same, and if we did make the mistake to identify them, we usually need to specify which kind of tangent vector we are talking about. Basically, we turn two non-ambiguous vector spaces into one ambiguous vector space, and I think it's not worth it.

Martin Brandenburg
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    I think you forgot the main reason why these canonical identifications don't get a name: c) when there are too many canonical isomorphisms it is a pain to come up with the notation that will write down everything explicitly. In fact, problems with not writing everything down are everywhere: every abelian group has the same name of the operation "+", everybody defines integral to be taken of a function and then without hesitation takes integral of a functional expression (same with derivatives)... the list can continue. – Matsmir Sep 02 '24 at 00:47
  • Hm? I never claimed that these need a name. – Martin Brandenburg Sep 02 '24 at 00:49
  • name in the broad sense: any symbol (or a string of symbols). – Matsmir Sep 02 '24 at 00:49
  • I think you misunderstood what we are talking about here. Any identification is done by providing a way of how this identification is done. By defining an isomorphism. I just advocate for not equating this isomorphism with the identity. It is not about giving that isomorphism any name. It just about not hiding it. In the proofs it will be used anyway. Your example with the $+$ is not related to the topic. – Martin Brandenburg Sep 02 '24 at 00:52
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    My comment mostly concerns the first paragraph of the answer where you give reasons why mathematicians accept that these isomorphisms are not written down and are replaced by identity. I claim that the most important reason is the simplification of notation. The example with the $+$ is just another place in mathematics where people give up some rigour to achieve simple notation. – Matsmir Sep 02 '24 at 00:56
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    I feel this a lot, and when I get confused with this things, I try to go back and write down all the isomorphisms explicitly; usually it is time-consuming, as no texts explicitly mention or follow these. – okabe rintarou Sep 02 '24 at 07:50
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    @Matsmir In most proof assistants (Lean 4 to be definitive), it is allowed to use + for different things, and no rigor is lost. I don't think any "rigor" beyond the level of computer checked proofs — where every detail is pre-programmed and mechanically checked — deserve to be called "rigor" at all. They are just pure tedium instead. – Trebor Sep 02 '24 at 08:22
  • @Trebor I don't know anything about Lean 4, so you should correct me if I am wrong. I think that allowing the same symbol to denote different objects could be possible for logic but it surely would impose some changes in the standard approach to it. Afaik it is not really a thing to take a variable in an expression and replace several occurences of it with one expression and other occurences with a different one (i.e. if you proved $\forall x P(x)$ and you want to apply it to some concrete $x$, that $x$ should be the same for all occurences). The $+$ in Lean 4 is more than a symbol, I guess. – Matsmir Sep 02 '24 at 09:28
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    Martin, I am very sympathetic to your point of view, but I'm just wondering how we can reconcile this with actual mathematical practice. For example, according to the standard constructions of $\mathbb N$ and $\mathbb Z$ in axiomatic set theory, it is not the case that $\mathbb N\subseteq\mathbb Z$. So should we stop doing this and instead write $\mathbb N\hookrightarrow\mathbb Z$? – Joe Sep 02 '24 at 20:23
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    More fundamentally, what do you think a symbol like $\mathbb N$ should mean? Should it denote a specific construction of the natural numbers, say $\omega$, or an arbitrary (yet fixed) ordered semiring satisfying the Peano axioms? What about, say, the polynomial algebra R[x]? – Joe Sep 02 '24 at 20:23
  • Yes exactly, there is a canonical map $\mathbb{N} \to \mathbb{Z}$. We don't need to remember its definition exactly and instead remember its characteristic property, it is the unique semiring homomorphism. (Actually, for every semiring $R$ there is a unique semiring homomorphism $\mathbb{N} \to R$.) More generally, for every semiring $R$ with associated Grothendieck ring $G(R)$ there is a canonical semiring homomorphism $R \to G(R)$. In this generality, it does not have to be injective! – Martin Brandenburg Sep 02 '24 at 20:25
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    As long as we have characteristic or universal properties, explicit constructions do not matter anymore once done, because the isomorphisms between the different models will be unique (when the category is chosen well enough). So I don't really care how you define $R[x]$, for me it's only important we talk about the free commutative $R$-algebra of rank $1$, i.e. the commutative $R$-algebra with $\hom(R[X],A) \cong U(A)$. If you have a different construction of the polynomial algebra, there will be unique isomorphism of $R$-algebras between them that preserves the variable. – Martin Brandenburg Sep 02 '24 at 20:27
  • Thank you for your response. By the way, I think I have another example of a genuine error that I've seen in a couple of places that comes from "identifications". When proving that every field $k$ has an algebraic closure, one often constructs a chain of extension fields $$k=k_1\hookrightarrow k_2\hookrightarrow k_3\hookrightarrow\cdots , ,$$with each $k_i$ canonically identified with a subfield of $k_{i+1}$. – Joe Sep 02 '24 at 20:38
  • A number of authors then try to take the union of the $k_i$, and prove that it is an algebraic closure of $k$. But this just doesn't make sense: $k_i$ is just not a subset of $k_{i+1}$ even if we are pretending that it is, and so there is no obvious field structure that can be put on the union of the $k_i$. There are two solutions, as far as I can tell: construct the $k_i$ so that the extensions are genuine inclusions, or take the direct limit of the $k_i$. – Joe Sep 02 '24 at 20:38
  • Forget the 1st "solution", it doesn't make any sense from a structural perspective. Also, the correct definition of a field extension is that it is just a field homomorphism. You don't need to change the $k_i \to k_{i+1}$. You just need to take the colimit, that's right. – Martin Brandenburg Sep 02 '24 at 22:00
  • Can you elaborate on the problem you have with the first solution? – Joe Sep 02 '24 at 22:56
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Here's the explanation I give when I teach differential geometry:

Choosing coordinates on a space is a multidimensional analogue of choosing units for a physical measurement such as length or temperature. Coordinates are extremely useful for doing physics and geometry.

On the other hand, a fundamental assumption of physics is that physical laws do not depend on the units or coordinates used to do measurements or calculations. Similarly, a fundamental assumption of geometry is that the geometric properties of space should not depend on any coordinates used to study the space.

There are two ways to show something is independent of the coordinates used. One is to use coordinates but then show that everything works consistently if you change the coordinates. The other is to not use coordinates at all.

The first place you see this is in modern linear algebra. An abstract vector space $V$ can be viewed as an abstract flat geometric space. A basis of $V$ defines an isomorphism $V\simeq\mathbb{R}^n$, i.e., coordinates on $V$. Geometric properties of abstract vector spaces and linear transformations should not depend on the basis (i.e., coordinates) used.

We know that there exists a linear isomorphism $L: V \rightarrow V^*$. If $(e_1, \dots, e_n)$ is a basis of $V$, then $L$ is uniquely determined by a basis $(f_1,\dots, f_n)$ of $V^*$ by setting $L(e_k) = f_k$. There are infinitely many such isomorphisms, all essentially equivalent to each other.

We could do the exact same thing with $V^{**}$. However, something cool happens here. There is an isomorphism \begin{align*} \phi: V &\rightarrow V^{**}, \end{align*} where for any $\ell \in V^*$, $$ (\phi(v))(\ell) = \ell(v). $$ This map does not depend on any basis of $V$ or $V^{**}$ because no basis is used in its definition. So it has a special status.

Note that this is true even for linear isomorphisms $V \rightarrow V$. There are many possible linear isomorphisms, but among them there is a special one, namely the identity map.

The story about the tangent space at a point on a manifold is a similar but more involved one.

user1551
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Deane
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  • I think you are missing the hypothesis that $V$ is finite-dimensional in your example. – Joe Sep 03 '24 at 01:34
  • @Joe, thanks. I should have said that. I never remember to do so, because in differential geometry we rarely study infinite dimensional manifolds. – Deane Sep 03 '24 at 01:43
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For $3)$, let $S\subset M$ be a submanifold of the same dimension as $M$, and $\iota:S\hookrightarrow M$ be the inclusion map. Since this is the $\textbf{inclusion map}$, for all $s\in S$, we have that $\iota(s)=s$. It follows that for all $v\in T_sS$, $$D_s\iota(v)=\frac{d}{dt}\Big|_{t=0}\iota(\gamma(t))=\frac{d}{dt}\Big|_{t=0}\gamma(t)=v$$ where $\gamma:I\rightarrow S$ is a smooth curve such that $\gamma(0)=s$, and $\dot{\gamma}(0)=v$. We thus have that $T_sM=D_s\iota(T_sS)=T_sS$. These are actual equalities, and this should really be expected as $S$ is a submanifold of the same dimension, and tangent vectors are derivations at a point which are in a sense "more" local than derivations on the whole space, i.e. they really only care about what a function is behaving like near that singular point.

Chris
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