Closed topological embedding of complex algebraic varieties into a smooth manifold

Question

In the book "Representation Theory and Complex Geometry” by Ginzburg-Chriss page 93 they claim that (the analytification of) every complex algebraic variety admits a closed embedding in a smooth manifold. They do not really provide a reference for this statement. Why is this true?

For quasi-projective varieties, the claim is fairly obvious. But it seems to me that they do not impose this restriction. Moreover, I would say that we need the variety to be separated for this but (I can live with that).

I think the most widespread definition includes separatedness. At least it is so in Wikipedia (they cite Hartshorne) — მამუკა ჯიბლაძე, May 27 '17 at 06:02
Yes, that's fine (I mean the claim would imply this anyways (my manifolds are Hausdorff at least)) — user110443, May 27 '17 at 06:06
As for references - they do refer to Goresky & Macpherson's "Intersection theory II" and to Rourke & Sanderson. It is true that it is not so easy to find relevant places there - G&M in turn refer to Mather's "Stratifications and mappings", while to use R&S I think one has first to ensure that one has a CW-complex structure — მამუკა ჯიბლაძე, May 27 '17 at 06:41
Yeah, I checked these two references they give but there's nothing. — user110443, May 27 '17 at 06:49
Goresky and Macpherson state it explicitly without proof. Did you look up in Mather? He must have it in detail I believe. — მამუკა ჯიბლაძე, May 27 '17 at 06:51
As for CW-complex structure: here was an answer somewhere with an argument showing that separated varieties have the homotopy type of a finite CW complex (also this is mentioned without proof in Ginzburg-Chriss and I could nowhere else find a reference). But even then I don't know how this helps for embedding questions. — user110443, May 27 '17 at 06:56
This (closed) question is about embedding CW-complexes, it is easy — მამუკა ჯიბლაძე, May 27 '17 at 07:20
Yeah, I know this, but it's just homotopy equivalent to a finite CW complex. — user110443, May 27 '17 at 07:49
Actually, it makes sense to add all of the above information into the body of your question, I believe this way it would become more clear what is the difficulty with it. — მამუკა ჯიბლაძე, May 27 '17 at 08:03
It was certainly not clear for at least one person. And with all the detailed references the question will be much better motivated, hence, according to my experience, will receive more eager attention from the potential answerers. — მამუკა ჯიბლაძე, May 27 '17 at 08:51
Also this answer sketches a proof for triangulability of any complex variety. — მამუკა ჯიბლაძე, May 27 '17 at 09:31
There's a PhD thesis "Triangulation of locally semi-algebraic spaces" by Kyle Roger Hofman proving triangulability for any variety. Still, it's not a finite simplicial complex I guess, so I don't know how to get this into a manifold. — user110443, May 27 '17 at 09:38
It is proved here (Theorem D on page 82 and some results above it) that a CW-complex embeds into ${\mathbf R}^n$ iff it is countable, locally compact, and has dimension $\leqslant n$. — მამუკა ჯიბლაძე, May 27 '17 at 09:58
Sounds promising actually. Any idea why the CW complex of a variety is countable? Locally compact of bounded dimension should be fine I guess. — user110443, May 27 '17 at 10:05
Okay, what about this argument (can some expert confirm or correct this?): let $X$ be a separated variety. By Nagata compactification, it's an open dense in a proper variety $\hat{X}$. By the answer here, $\hat{X}$ admits a finite triangulation. Hence, $\hat{X}$ can be embedded in a smooth manifold $M$. Then $X$ is locally closed in $M$, so $X = U \cap Z$ for an open $U$ and a closed $Z$ in $M$. Hence, $X$ is closed in an open subset $U$ of $M$, and this is again a smooth manifold. — user110443, May 27 '17 at 11:13
Why not post it as an answer? This way it is more chance that a proper expert will see it :P — მამუკა ჯიბლაძე, May 27 '17 at 12:27
The statement should be easier than that. Use that every variety is covered by finitely many affines. Affines embed in A^n. Thenuse the normal whitney embedding arguemnt to embed them in — Atticus Christensen, May 28 '17 at 13:47
You mean the statement or its proof? Concerning the latter, I agree. However, along your lines I don't see how to globally glue this into a closed set of a single manifold. — user110443, May 28 '17 at 23:58

user110443 · Answer 1 · 2017-05-30T04:09:36.700

It took me ages to find this reference (and I actually found it in this post). There is a theorem by Acquistapace–Broglia–Tognoli in An embedding theorem for real analytic spaces stating:

Theorem. Let $X$ be a paracompact connected $n$-dimensional analytic space and suppose that $q := \sup_{x \in X} \dim T_x X < \infty$. Then $X$ admits a closed $C^\omega$-immersion into $\mathbb{R}^{n+q}$.

This is a generalization of Grauert's embedding theorem for real analytic manifolds. From this we obtain:

Corollary. Let $X$ be a separated complex algebraic variety. Then $X^{\mathrm{an}}$ admits a topological closed embedding into $\mathbb{R}^N$ for some $N$.

Proof. $X^{\mathrm{an}}$ is a paracompact complex (thus also real) analytic space with finitely many connected components and with bounded tangent space dimensions. Now, apply the theorem.

Remark 1. The only thing that surprises me is why the ABT paper has only one citation on MathSciNet since 1979; I don't know if there's another source for the above theorem (I hope it's right actually).

Remark 2. I don't know if there isn't a simpler argument.

Closed topological embedding of complex algebraic varieties into a smooth manifold

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