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I first got concious of the notion of normal varieties around 3 years ago and despite the fact that by now I can manipulate with it a bit, this notion still puzzles me a lot. One thing that strikes me is that the definition of normality is so entirely algebraic.

From my common sense understanding the notion of normal varieties restricts the class of spaces that we consider to more-less reasonable ones. It looks to me that this definition is analogous to the definition of pseudo-manifold. At least the obvious similarity is that in both cases the set of non-singular points is connected.

Normality pops up everywhere and its definition is very short. But it is hard for me to imagine that a differential topologist or differential geometer could come up with such a definition. Why is the notion of normatilty is so omnipresent? What is "geometric" meaning of normality?

Maybe a more concrete question would be like this. Suppose $X$ is an irreducible algebraic subvariety in $\mathbb C^n$ with singularities in co-dimension $2$. Can one somehow looking on singularities, their stratification and the way $X$ lies in $\mathbb C^n$ say if it is normal or not?

Added. Who was the person who invented this notion?

I would like to thank everybody for useful comments and links.

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    $\begingroup$ I don't know how helpful this is for you purposes, but there is the Serre criterion for normality: That $X$ satisfies your condtion (that is, regular in codimension 1, usually written $R_1$) and the more technical condition $S_2$, which can be difficult to verify. See also Hailong Dao's answer to this question: mathoverflow.net/questions/60097/… $\endgroup$
    – J.C. Ottem
    Commented Oct 11, 2012 at 17:48
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    $\begingroup$ Maybe you want to check out Sandor Kovacs answer to this question. mathoverflow.net/questions/35736/… $\endgroup$
    – Thomas Kahle
    Commented Oct 11, 2012 at 18:54
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    $\begingroup$ mumford emphasizes that normal implies locally unibranch. i like the criterion that a variety is normal iff any surjective finite birational map onto it is an isomorphism. $\endgroup$
    – roy smith
    Commented Oct 11, 2012 at 23:19
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    $\begingroup$ This might be a silly comment but it seems to me that normality is a way to get an algebraic version Hartog's principle. $\endgroup$
    – DamienC
    Commented Oct 12, 2012 at 19:01
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    $\begingroup$ normal varieties were introduced by Zariski, in a paper in the Amer. Journal of Math, vol. 61, 1939,p.249ff? but announced in an earlier paper in 1937. $\endgroup$
    – roy smith
    Commented Oct 12, 2012 at 20:11

5 Answers 5

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This is basically the same as roy smith's excellent comment, but I'd like to put a slightly different spin on it.

A normal variety is a variety that has no undue gluing of subvarieties or tangent spaces.

Let me explain what I mean by gluing. Given a variety $X$, a closed sub-scheme $Y \subseteq X$ and a finite (even surjective) map $Y \to Z$, you can glue $X$ and $Z$ along $Y$ (identifying points and tangent information). This is the pushout of the diagram $X \leftarrow Y \rightarrow Z$.

You might not always get a scheme (although you do in the affine case) but you always get an algebraic space. In the affine case, this just corresponds to the pullback in the category of rings.

Example 1: $X = \mathbb{A}^1$ glued to $Z = \bullet$ (one point) along $Y = \bullet, \bullet$ (two points) is a nodal curve.

Example 2: $X = \mathbb{A}^1$ glued to $Z = \bullet$ (one point) along $Y = \star = \text{Spec } k[x]/x^2$ a fuzzy point gives you a cuspidal curve.

Example 3: $X = \mathbb{A}^2$ glued to $Z = \mathbb{A}^1$ along one of the axes $Y = \mathbb{A}^1$ via the map $Y \to Z$ corresponding to $k[t^2] \subseteq k[t]$ gives you the pinch point / Whitney's umbrella = $\text{Spec } k[x^2, xy, y]$.

If I recall correctly, all non-normal varieties $W$ come about this way for some appropriate choice of normal $X$ (the normalization of $W$) and $Y$ and $Z$ (NOT UNIQUE). Roughly speaking, if you are given $W$ and want to construct $X, Y, Z$, do the following: Let $X$ be the normalization, let $Z$ be some sufficiently deep thickening of the non-normal locus of $X$ and let $Y$ be some appropriate pre-image scheme of $Z$ in $X$.

Edit: There is a proof available now HERE

Assuming this is true, you can see that all non-normal things are non-normal because they either have some points identified (as in 1 or 3) or some tangent space information killed / collapsed (as in example 2), or some combination of the two.

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    $\begingroup$ very nice job. much more pregnant and also more clear and precise than my comment. this puts flesh on it and cries out for exploration. $\endgroup$
    – roy smith
    Commented Oct 13, 2012 at 0:03
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    $\begingroup$ Here is an example I find hard to visualize fully geometrically by the method I callee "geometric": take a smooth rational quartic curve in P^3 and let X be the cone over it in P^4. This seems to be a standard example of a non normal surface satisfying R1 but not S2. One can use geometry to give a finite birational map to X (I hope), (by projecting the cone in P^5 over a rational quartic curve in P^4 onto X), but is there a "geometric" way to see this map is non trivial? My feeling now is that S2, i.e. depth, is hard to make fully geometric. $\endgroup$
    – roy smith
    Commented Oct 24, 2012 at 2:50
  • $\begingroup$ ok, i'm going to say that i am back to mumford's point, i.e. the only finite birational maps that are (to me) obviously not isomorphisms are those that are not bijective. so the only case where i can really see geometrically that a variety is not normal is one that is not unibranch. then its normalization is not bijective. so this is a little larger class than "not R1", but not fully the class of "not S2". i.e. I can really see geometrically that a non unibranch variety is not normal, but it is not so easy to see geometrically that a variety is not S2. $\endgroup$
    – roy smith
    Commented Oct 24, 2012 at 6:08
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    $\begingroup$ So, the other way to break non-S2-ness, besides gluing points, is to kill tangent information at a point. That is what's going on with the first example (cone over the quartic rational curve in $\mathbb{P}^3$) although I must admit, I don't have a good way to visualize that example But generally, the easiest way to make a non-normal graded ring is to kill some low degree terms in a normal graded ring. This will make something that is not S2 (unless the graded ring has dimension $\leq 1$). $\endgroup$
    – Karl Schwede
    Commented Oct 24, 2012 at 11:33
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    $\begingroup$ @KarlSchwede: Do you know a reference to the fact that any non-normal variety can be gotten by this gluing construction? $\endgroup$
    – Joshua Grochow
    Commented Oct 23, 2013 at 1:38
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As J. C. Ottern has pointed out, Serre's criterion gives a clue. Normality is equivalent to $R_1 + S_2$. The interpretation of $R_1$ is easy: regular in codimension 1, the singular locus has codimension at least two, it is small, e.g. for a surface this means that the singular points are isolated. The interpretation of $S_2$ is the "extension property", every algebraic function defined on a open set whose complement is of codimension at least two, extends to all the variety. The proof of this fact is essentially contained in EGA IV$_2$ $\S$5.10.

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    $\begingroup$ Leo, I looked up the reference in numdam. Unfortunately this language for the moment is superior to what I can possibly understand. But thank you for the answer and the reference! $\endgroup$
    – aglearner
    Commented Oct 11, 2012 at 23:29
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    $\begingroup$ See a explanation here: mathoverflow.net/questions/45347/… $\endgroup$
    – temp
    Commented Oct 12, 2012 at 1:24
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At least in the case of complex algebraic varieties one can give a nice topological interpretation of the normality condition. Let us consider $V$ a complex algebraic variety, then its complex points $V(\mathbb{C})$ has the structure of a stratified pseudomanifold.

Let me recall that a stratified pseudomanifold $X$ is a filtered topological space $$X_0\subset\ldots \subset X_n$$ such that each stratum, i.e. a connected component of $X_i-X_{i-1}$ is a manifold of dimension $i$ and such that $X_{n-1}=X_{n-2}$ and such that the regular part $X_n-X_{n-2}$ is dense in $X$. Together with a local condition: the existence of conical charts.

Thus $V(\mathbb{C})$ comes equipped with such a geometric structure. In the setting of stratified pseudomanifold one has a notion of normal pseudomanifold and normalization is a fundamental concept in intersection homology. A pseudomanifold $X$ of dimension $n$ is said to be normal if for every point $x\in X$ the local homology group $H_n(X,X-x,\mathbb{Z})$ is isomorphic to $\mathbb{Z}$. Notice that a homological manifold is normal. Using Zariski’s Main Theorem, one can prove that a normal complex algebraic variety is a normal pseudomanifold.

If you consider a triangulation $T$ of $X$ ($dim(X)=n$) then you can also prove that $X$ is normal if and only if the link of eack simplex in the $n-2$-skeleton of $T$ is connected. This is proved in Goresky, MacPherson "Intersection Homology theory" (Topology Vol. 19 (1980)). In this paper the authors also explains how to build normalization topologically and how topological normalization satisfies a universal property. In the case of $V(\mathbb{C})$ its topological normalization in the sense of Goresky-MacPherson is homeomorphic to $V'(\mathbb{C})$ where $V'$ is the algebraic normalization of $V$.

Thus topologically normality corresponds to the connectivity of the links, the link of a point in an $n$-dimensional manifold being a $n-1$ sphere we see that topological normalization is the very first step to desingularization of stratified pseudomanifolds.

Here are two examples:

1) The pinched torus is not normal. It is a complex projective curve $C$ of equation $x^3+y^3=xyz$ in homogeneous coordinates $[x:y:z]$. It has a unique singular point $[0:0:1]$ and the link of this point $p$ is homeomorphic to two circles (we have $H_2(C,C-p;\mathbb{Z})\cong \mathbb{Z}\oplus \mathbb{Z}$).

2) The quadric cone is normal. It is an algebraic surface $S$ of equation $x^2+y^2+z^2=0$ in $\mathbb{P}^3(\mathbb{C})$ in homogeneous coordinates $[x:y:z:w]$ it has a unique singular point $[0:0:0:1]$. We notice that this space is homeomorphic to the Thom space of the tangent bundle of the $2$-sphere $S^2$. This remark gives a homeomorphism between the link of the singular point and the unit sphere bundle of the tangent bundle of $S^2$ which is connected (we get that $S$ is topologically normal).

Historicaly these two examples were important for our understanding of the failure of Poincaré duality for singular spaces, they appear in Zeeman's thesis:

E. C. Zeeman, "Dihomology III. A generalization of the Poincaré duality for manifolds", Proc. London Math. Soc. (3), 13 (1963), 155-183.

and also in McCrory's thesis:

C. McCrory, "Poincaré duality in spaces with singularities", Ph.D. thesis (Brandeis University, 1972)

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    $\begingroup$ I'm confused about the case of a cuspidal curve. Isn't the link of the singularity of $y^2=x^3$ the trefoil knot? This is not normal, but the link seems to be connected. What am I missing? $\endgroup$
    – Jim Bryan
    Commented Dec 28, 2012 at 16:56
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    $\begingroup$ I should have said that normal in the topological sense does not imply normal in the algebraic sense. The only thing we can say is that the cuspidal curve is topologically normal thus homeomorphic to its algebraic normalization. $\endgroup$
    – David C
    Commented Dec 28, 2012 at 17:18
  • $\begingroup$ Dear David, thank you for this asnwer! $\endgroup$
    – aglearner
    Commented Dec 30, 2012 at 12:21
  • $\begingroup$ @DavidC Do you know if perhaps "topologically normal" is equivalent to "Locally Unibranch" for reduced varieties over $\mathbb{C}$? $\endgroup$
    – Saal Hardali
    Commented Apr 26, 2018 at 17:18
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An excellent non-algebraic meaning (using analysis) of normality is found in Kollar's article in the Bulletin of AMS (1987).

Restrict to irreducible varieties $X$ so we can talk of function fields. A point $x_0\in X$ is considered normal whenever a rational function exhibits decent behaviour in a neighbourhood of $x_0$ then it finds a place in the local ring of $X$ at $x_0$.

Decent behaviour here is: If $f\in K(X)$ and if $|f(x)|$ remains a bounded function as $x$ approaches $x_0$ by paths lying in $X$, then $x_0$ should be good enough to admit $f$ in its local ring.

This survey article of Kollar is about Mori's Fields-medal winning work on 3-folds. But it starts from the scratch defining what an algebraic variety is. It is a great source to learn the meanings of fundamentals objects of algebraic geometry. (for example Kollar explains why we have to deal with line bundles when we study projective varieties).

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    $\begingroup$ Dear P Vanchinathan, thank you very much for recommending this article! $\endgroup$
    – aglearner
    Commented Dec 30, 2012 at 12:11
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Regarding the question "Who was the person who invented this notion?", a paper of H. T. Muhly provides interesting background (as well as a geometric interpretation) for projectively normal:

In the terminology of the Italian School an algebraic variety is called "normal" if its system of hyperplane sections is complete. O. Zariski applies the term "normal" to an algebraic variety whose associated ring of homogeneous coordinates is integrally closed. The two concepts are not equivalent. Zariski refers to a variety which satisfies the former condition as "normal in the geometric sense" and to one which satisfies the latter condition as "normal in the arithmetic sense".

(...)

The object of this note is to characterize geometrically those algebraic varieties which are normal in the arithmetic sense. To this end we propose the following theorem: A necessary and sufficient condition that the $r$-dimensional algebraic variety $V_r$ be normal in its ambient projective space $P_n$ is that for every integer $m$ the linear system cut out on $V_r$ by the hypersurfaces of order $m$ in $P_n$ be complete.

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  • $\begingroup$ Dear Francois, I think the last quoted paragraph is about characterizing projectively normal varieties. Regards, Matthew $\endgroup$
    – Emerton
    Commented Dec 27, 2012 at 21:38
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    $\begingroup$ Matthew -- Thanks; you are quite right and I have edited accordingly. (To me the main interest here was the statement that normality has origins earlier than Zariski.) $\endgroup$
    – Francois Ziegler
    Commented Dec 27, 2012 at 22:22
  • $\begingroup$ Dear Francois, it is ineed very interesting to know that Italians were already thinking of normality :) $\endgroup$
    – aglearner
    Commented Dec 30, 2012 at 12:12

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