MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know in differential geometry, given a $C^k$ manifold for $k>1$, the tangent space at a point in this manifold is parametrized by curves passing through this point modulo certain equivalence relation. The tangent space is given by the velocity vectors to these curves at this point.

Intuitively, I would think this should be true even for Zariski topology in the case $X$ is a smooth projective variety. In particular, we could take a tangent vector $t$ and associate to this the subscheme $Y_t$ of $X$ obtained by intersecting all some subschemes of $X$ which has $t$ as an element in its tangent space. The question is whether we can somehow ensure that $Y_t$ is $1$-dimensional?

share|cite|improve this question
In the smooth category, the intersection you mention would only be the point (0-dimensional) provided your manifold has dimension $> 1$. – Fernando Muro Apr 27 '13 at 1:41
The only common approach to tangent spaces I know is the usual definition: the derivations of the local ring at the point with coefficients in the residue field. – Fernando Muro Apr 27 '13 at 1:45
The tangent space of a locally ringed space $X$ at a point $x$ is $(\mathfrak{m}_x/\mathfrak{m}_x^2)^*$, where the dual is taken over the residue field $\kappa(x)$. But probably this doesn't answer your question, since you want to use curves into $X$? If $X$ is a scheme over $k$, the replacement for the "infinitesimal interval" is the spectrum of the ring of dual numbers $k[t]/t^2$. – Martin Brandenburg Apr 27 '13 at 9:09
Can't you work out an example before asking a question? How can you think that the intersection of all curves with a given tangent vector is 1-dimensional? – Angelo Apr 27 '13 at 13:22
I don't think rudeness is really necessary. One of the things that I really like about this site is that one can ask questions (about things that might be frighteningly "obvious" to others) without being verbally cut-down. Though this may the norm in certain realms of academia, there's no reason for it to be. – Jon Beardsley May 2 '13 at 2:06
up vote 10 down vote accepted

If p is a nonsingular point, then we can define a tangent vector as an equivalence class of (nonsingular) curves, just as in the differentiable case. In fact, being nonsingular is equivalent to every tangent vector being the speed* of a curve. In this case, the intersection of all curves in the equivalence class is a zero-dimensional closed subscheme with a one-dimensional tangent space, isomorphic to the spectrum of the dual numbers described above by Martin Brandenburg.

*Not sure this is the right term, I didn't learn differential topology in English.

Warning: I'm thinking of schemes of finite type over an algebraically closed field K, and by point I mean a K-valued point.

share|cite|improve this answer
@Barbara: Thanks. This is exactly what I was asking. Do you have a reference in mind for this? – Jana Apr 27 '13 at 16:18
Unfortunately I don't. If it's useful I could add more details. – Barbara Apr 27 '13 at 19:58
@Barbara: The word you're probably looking for is "velocity", not "speed". From wikipedia: "velocity is the rate of change of the position of an object, equivalent to a specification of its speed and direction of motion." In English at least, I think "velocity" is meant to refer to the vector and "speed" is meant to refer to the scalar (the magnitude of the vector). – Kevin H. Lin May 2 '13 at 8:51

One nice way to say this in the language of schemes is that the tangent space of a (say complex) variety $X$ at a point $p$ is the set of morphisms from the scheme $Spec(\mathbb{C}[\epsilon]/(\epsilon^2))$ to $X$ which send the closed point to $p$. In other words, we are looking at paths in $X$, passing through $p$, but we only consider the paths up to 1st order near $p$.

Chapter 12.1 of Vakil's notes gives a detailed discussion of the Zariski tangent space. The above fact appears as exercise 12.1.I

share|cite|improve this answer

I would say that it is easier to view the things the other way round and to transport in differential geometry the algebraic-geometry definition of differentials. This is the one which is alluded to by Martin Brandenburg when he writes the tangent space $T_xX$ as the dual of $\mathfrak m_x/\mathfrak m_x^2$.

What the Taylor expansion of $f$ says is that a function $f$ is equal to its value at $x$ plus its differential plus something which has order $2$. Moreover, a function which has order $2$ at $x$ can be written as a linear combination of products of two functions vanishing at $x$.

This leads to the definition: the space of differentials at $x$ is the quotient of the ideal $\mathfrak m_x$ of the ring of $C^\infty$-functions vanishing at $x$ modulo its square. Then the differential of $f$ (satisfying $f(x)=0$) is just `$f$ up to second order terms''.

share|cite|improve this answer

$T_xX$ as the dual of $\mathfrak m_x/\mathfrak m_x^2$ corresponds to tangent vectors acting as derivations over the point evaluation at $x$ on the function algebra. $T_xX$ as equivence classes of curves through $x$ correspond to velocity vectors at $x$, so they help kinematic visualization. On (at least $C^1$-) finite dimensional manifolds (including smooth points of varieties) both notions coincide. In singular points there are many more "operational" than "kinematic" tangent vectors.

  • But even on an infinite dimensional Hilbert space the same is true: see 28.3 and 28.4 of here.
share|cite|improve this answer

When the variety $X$ is affine $n$-space and you take the curves to be maps from $\mathbb{A}^1$ to $X$, then the differential geometry description of the tangent space works. In the general case there is a map from an open neighbourhood of the point $P$ in $X$ to affine space sending $P$ to $0$ and inducing an isomorphism from tangent space at $P$ to that at $0$ (this just says that $X$ is locally isomorphic to affine space in the etale topology). Unfortunately, the maps from $\mathbb{A}^1$ to $\mathbb{A}^n$ need not lift to $X$, so you need to allow maps from smooth curves into $X$ (pull-backs of the covering $X\to \mathbb{A}^n$ by the maps from $\mathbb{A}^1$ into $\mathbb{A}^n$). So the tangent space at $P$ can be identified with certain equivalence classes of maps from smooth curves into $X$.

share|cite|improve this answer

If you want to see that every tangent line to an r dimensional projective variety is the tangent space to a curve through a smooth point p, just take r-1 functions in the local ring whose differentials cut out precisely the tangent line. Then the basic implicit function theorem type results on pages 7-9 of Mumford's "yellow book", Complex projective varieties, show that these functions define a curve through p with the given tangent line as tangent space. So the tangent space at a smooth point p, is the union of the tangent spaces of all curves through p and smooth at p.

So you get almost your point of view but replacing viewing the tangent vectors as velocity vectors to parametrized curves, by viewing the tangent lines as tangent spaces to implicitly given curves.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.