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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?

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This answer: mathoverflow.net/questions/14446/normal-bundle-to-a-curve-in-p2/… discusses a related question, and may be helpful. –  Emerton Feb 7 '11 at 4:51
    
@Emerton: Thanks a lot! How can we relate this to "higher infinitesimal"? –  Brian Feb 7 '11 at 14:51
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2 Answers 2

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 $\mathscr I/\mathscr I^2$.

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Thanks a lot for your answer! How does this apply to higher infinitesimal? –  Brian Feb 7 '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 7 '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 8 '11 at 1:21
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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 8 '11 at 5:46
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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.

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