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Let $f:X\to Y$ be a morphism between schemes. To construct the relative sheaf of differentials on $X$ (relative to $Y$), we first consider the diagonal map $\Delta: X \to X\times_Y X$ and then define $\Omega_{X/Y} = \Delta^{-1} \mathscr{I}/\mathscr{I}^2$ where $\mathscr{I}$ is the sheaf representing the immersion $X\to X\times_Y X$ (it's the kernel of $\mathscr{O}_{X\times_Y X} \to \Delta_* \mathscr{O}_X$.

Algebraically, this works out fine, due to the theory of abstract Kahler derivation defined on algebras. Is there a way to actually see the motivation behind this?

Moreover, what's the analog in higher infinitesimal approximation (instead of just 1st order one given by the differentials)? What's the (say, "analytic") insight behind the relationship between higher infinitesimal and higher diagonal?

Brian
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  • This answer: http://mathoverflow.net/questions/14446/normal-bundle-to-a-curve-in-p2/14470#14470 discusses a related question, and may be helpful. – Emerton Feb 07 '11 at 04:51
  • @Emerton: Thanks a lot! How can we relate this to "higher infinitesimal"? – Brian Feb 07 '11 at 14:51

2 Answers2

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Since we are talking about motivation let's assume that $X$ is smooth over $Y$. Also by slight abuse of notation let $\Delta$ denote the image of the diagonal map in $X\times_Y X$.

For a submanifold of a manifold one has a short exact sequence connecting the tangent bundle of the ambient manifold restricted to the submanifold, the tangent bundle of the submanifold and the normal bundle of that submanifold in the ambient manifold. The geometric explanation to why the definition of the cotangent sheaf via the conormal bundle of the diagonal in the self-product is the right one is that the normal bundle of the diagonal is isomorphic to its tangent bundle and the (co)normal bundle can be defined without the tangent bundle, so the tangent bundle may be defined as the normal bundle for this particular embedding.

In algebraic geometry we usually prefer the dual version involving the cotangent bundles (and of course, in general cotangent sheaves), so let's work with that. Write down the relevant short exact sequence for $\Delta\subset X\times_Y X$:

$$ 0 \to \mathscr I/\mathscr I^2 \to \Omega_{X\times_Y X/Y}\otimes \mathscr O_{\Delta} \to \Omega_{\Delta/Y} \to 0. $$

Observe that $\Omega_{X\times_Y X/Y}\simeq p_1^*\Omega_{X/Y}\oplus p_2^*\Omega_{X/Y}$ and hence $\Omega_{X\times_Y X/Y}\otimes \mathscr O_{\Delta} \simeq \Omega_{\Delta/Y}\oplus \Omega_{\Delta/Y}$. In fact, the natural morphism $\Omega_{X\times_Y X/Y}\otimes \mathscr O_{\Delta} \to \Omega_{\Delta/Y}$ in the above short exact sequence may be identified with either projection to one of the direct summands since restricting either projections to the diagonal induces an isomorphism, or perhaps even better to say that the diagonal is a section of either projection. It follows that $\mathscr I/\mathscr I^2\simeq \Omega_{\Delta/Y}$. Since the diagonal morphism is an isomorphism between $X$ and $\Delta$, it is clear that whatever way we define $\Omega_{X/Y}$, it has to be isomorphic to the pull-back of $\mathscr I/\mathscr I^2$.

Sándor Kovács
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  • Thanks a lot for your answer! How does this apply to higher infinitesimal? – Brian Feb 07 '11 at 14:49
  • Brian, if say $Y$ is an algebraically closed field and $X$ is smooth over $Y$, then $\mathscr I^n/\mathscr I^{n+1}$ is isomorphic to the $n^{\rm th}$-symmetric power of $\mathscr I/\mathscr I^2$, so you could identify them with pluriforms. – Sándor Kovács Feb 07 '11 at 17:35
  • Thanks! This is still a bit too "algebraic". What is the, say, "analytic" idea behind it, if there is any? Thanks! – Brian Feb 08 '11 at 01:21
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    OK, let's say it "analytically": $X$ is a manifold, $Y$ is a point, so $X\times_YX$ is just $X\times X$. The order $n$ infinitesimal neighborhood of the diagonal in $X\times X$ (which is isomorphic to $X$) can be identified with the $n^{\rm th}$ symmetric power of the tangent bundle of $X$. – Sándor Kovács Feb 08 '11 at 05:46
  • Dear @SándorKovács, how is the $n^\text{th}$ infinitesimal neighborhood isomorphic to the $n^\text{th}$ symmetric power of the tangent bundle? This suggests $n$-jets are "globally polynomial" and seems to contradict this comment. What am I missing? – Arrow Jan 08 '21 at 16:16
  • @Arrow, I'm not entirely sure what you are asking. The symmetric power of a bundle is a local construction, it does not guarantee that the global sections of the $n^\text{th}$ symmetric power is the $n^\text{th}$ symmetric power of the global sections of the original bundle. I also don't see how this relates to the linked comments. Can you try to explain what you mean? – Sándor Kovács Jan 08 '21 at 20:05
  • Dear @SándorKovács, the phrase "globally polynomial" was poor chosen. I am asking about the seeming contradiction between your assertion and the comment by Tom Goodwillie asserting that jet bundles are not obtained by applying a functor fiberwise to the tangent bundle. – Arrow Jan 08 '21 at 20:40
  • @Arrow, OK, I think I see the issue. The n-jets don't correspond to the $n^\text{th}$ infinitesimal neighbourhood. The latter is the "difference" between n-jets and (n-1)-jets. More precisely the $n^\text{th}$ infinitesimal neighbourhood corresponds to the kernel of the restriction map from n-jets to (n-1)-jets. If you look at Goodwillie's previous comment, then he says exactly what I am saying (well, for n=2 at least). Does this resolve your worry? – Sándor Kovács Jan 08 '21 at 20:52
  • Dear @SándorKovács, I think I agree the kernel of the map "omitting $n^\text{th}$ derivatives" is the $n^\text{th}$ symmetric power of the tangent bundle, since in coordinates the kernel consists of the degree $n$ homogeneous polynomials. However, I think the $n$-jets do correspond to the $n^\text{th}$ infinitesimal neighborhood. – Arrow Jan 08 '21 at 21:03
  • @Arrow, OK, I think you are right and I hadn't realized until now, but apparently I have been using the term "$n^\text{th}$ infinitesimal neighbourhood" incorrectly in the above comment. Then just ran with it... Anyway, I meant $I^n/I^{n+1}$, not $\mathscr O/I^{n+1}$. I don't know if that has a name... – Sándor Kovács Jan 09 '21 at 00:28
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I would like to add another answer to this old question. Consider the case $X = Spec(A)$, $Y = Spec(R)$. Just to fix ideas, suppose that $A = R[T]$. If $f\in A$ and $a_0\in R$, one can consider the Taylor expansion of $f$ around $a_0$:

$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$

Now there is no reason why we should take a rational point $a_0 : R[T] \to R$ and in fact we can consider the Taylor expansion around an arbitrary $S$-valued point $a_0 : R[T]\to S$. The Taylor expansion will then be naturally an element of $S\otimes_R R[T]$. Taking the universal point $S = R[T_0]$, $a_0 = T_0$, we see that the ``universal Taylor expansion'' of $f$ is $$ f(T) = \sum_i\frac{f^{(i)}(T_0)}{i!}\cdot (T-T_0)^i\in R[T_0,T]. $$ If we write $R[T_0,T] = R[T]\otimes_R R[T]$, then we rewrite the above as $$ 1\otimes f(T) = \sum_i\left(\frac{f^{(i)}(T)}{i!}\otimes 1\right)\cdot(1\otimes T-T\otimes 1)^i $$ Looking mod $(1\otimes T-T\otimes 1)^2$ we get: $$ 1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2} $$

Now, in this particular case, $I =\ker(A\otimes_R A\to A)$ is generated by $1\otimes T-T\otimes 1$. Hence we see that $I/I^2$ is simply the space of linear terms of Taylor expansions and the canonical map $d : A\to I/I^2$ is simply sending a function $f\in A$ to the linear term in its Taylor series. Note that $1\otimes T-T\otimes 1$ is usually denoted by $dT$.

This also explains nicely what happens in higher degree. We can introduce the algebras $P^n = (A\otimes_R A)/I^{n+1}=R[T_0,T]/(T-T_0)^{n+1}$, the ring of Taylor expansions of degree $\leq n$ where the terms of degree at most $n$ of the Taylor expansion live. There is a natural map $d^n : A\to P^n$, sending $a$ to $1\otimes a$ which is simply sending $a$ to its Taylor expansion.

This explanation works exactly the same if $A/R$ is smooth (instead of $A = R[T]$), because locally on $A$ there is an etale map $F\to A$ where $F$ is a polynomial $R$-algebra and this map induces an isomorphism on $I/I^2$ and $P^n$ more generally.

Nicolás
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  • Dear Nicolás, isn't it customary to denote by $\mathrm df$ the homogeneous first order part of the Taylor expansion of $f$, i.e $\mathrm df=f^\prime(t)\mathrm dt$? The map you call $\mathrm d^n$ is more like the $n$-jet map, no? – Arrow Jan 02 '21 at 09:26
  • Possible, I don't remember now what is the right notation for this map. Thanks! – Nicolás Jan 03 '21 at 16:40
  • Just thought I'd point out that the notation $\mathrm d^n$ is used by Grothendieck in EGA IV, part 4, definition 16.3.6. – Arrow Jan 30 '21 at 11:17