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Let $A$ be a commutative noetherian ring, and $P$ a projective $A$ module with $rank(P)=n$. I know that $\wedge^nP \simeq L$ for some rank 1 projective $A$-module, $L$; but I'm not sure of how to think of this $L$. I understand that the top exterior power of a vector space with basis $\{e_1, e_2, \ldots, e_n\}$ is 1 dimensional with $e_1 \wedge e_2 \wedge \ldots \wedge e_n$ as its basis. But how does one expand this concept to a projective module where there is only a local basis?

As a more specific question, let $A$ be the coordinate ring of the real 2-sphere, and $P$ the rank 2 projective module corresponding to the tangent space of $A$. $P$ is indecomposable, and there are no non-trivial rank 1 projective modules over $A$. So, $\wedge^2P \simeq A$, but I can't justify this any other way than by pigeonholing it. My main problem is that $P$ doesn't have a global basis, thus negating (as far as I can tell) the idea of taking the wedge product of basis elements.

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If there's a local basis, then there's a local rank. Isn't it enough to show that the local rank is equal to one everywhere? (Also, the ring in your specific question fails to be noetherian, unless you mean to take $A$ to be entire functions on the Riemann sphere or something equally restrictive...) –  Zhen Lin Mar 19 '12 at 7:28
    
Specifically: Let $A=\mathbb{R}[X,Y,Z]/(X^2+Y^2+Z^2-1)$,(use $x,y,z$ to represent the images of $X,Y,Z$ in $A$), then let $\phi:A^3 \rightarrow A$ given by multiplication by $x$ in the first coordinate, $y$ in the second, and $z$ in the third (this is a surjective map since $\phi(x,y,z)=1$), and finally $P=ker(\phi)$. $A$ and $P$ are as I describe above. –  Andrew Parker Mar 19 '12 at 11:56
    
And regarding local rank one implies globally rank one: a height $n$ local complete intersection may be locally $n$-generated, but not globally so. Moreover, I'm interested in what is really going on with wedge product when there are no globally defined basis elements - not the specific calculation of rank. –  Andrew Parker Mar 19 '12 at 12:11
    
Ah, so you're trying to do algebraic geometry rather than differential geometry. You might want to retag. Your construction is interesting. Assuming that everything can be interpreted geometrically – and this is troublesome because $\mathbb{R}$ is not algebraically closed – then one can argue by differential geometry that $L$ should be the free module of rank $1$: the top exterior power of the cotangent bundle has a global generator, namely $\mathrm{d}x \wedge \mathrm{d}y + \mathrm{d}y \wedge \mathrm{d}z + \mathrm{d}z \wedge \mathrm{d}x$, and the dual of a trivial bundle is a trivial bundle. –  Zhen Lin Mar 19 '12 at 12:20
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@Andrew: Look for "determinant of a vector bundle" in the literature. –  Martin Brandenburg Mar 19 '12 at 19:01

1 Answer 1

up vote 3 down vote accepted

As requested, I am explaining the differential geometry point of view; I am assuming that you will be able to work out the commutative algebra on your own.


It is a well-known fact in differential geometry that a smooth manifold is orientable if and only if the top exterior power of the cotangent bundle has a global generator. The sphere is clearly orientable, so the vector bundle of differential $2$-forms must have a global generator – so we just need to find one.

First of all, let $M = \{ (x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \}$. Then, by abuse of notation, we have smooth functions $x, y, z : M \to \mathbb{R}$. The tangent bundle is the manifold $$T M = \{ (x, y, z, u, v, w) \in \mathbb{R}^6 : x^2 + y^2 + z^2 = 1, x u + v y + z w = 0 \}$$ equipped with the evident projection down to $M$. The cotangent bundle is the fibrewise dual of this vector bundle, denoted by $T^* M$ or $\Omega^1_M$. The exterior derivative is a map of sheaves $\mathrm{d} : C^\infty_M \to \Omega^1_M$ such that $\langle \mathrm{d} f , X \rangle$ is the directional derivative of $f$ along the vector field $X$. In particular, we have global sections $\mathrm{d}x, \mathrm{d}y, \mathrm{d}z : M \to \Omega^1_M$, and these satisfy the equation $$x \, \mathrm{d} x + y \, \mathrm{d} y + z \, \mathrm{d} z = 0$$

Since $M$ is an embedded submanifold of $\mathbb{R}^3$, it inherits a Riemannian metric; it is the global section $g : M \to T M \otimes T M$ given by $$g = \mathrm{d}x \otimes \mathrm{d}x + \mathrm{d}y \otimes \mathrm{d}y + \mathrm{d}z \otimes \mathrm{d}z$$ By non-degeneracy, this induces an isomorphism of vector bundles $T M \cong T^* M$, so to show that $\Lambda^2 T M$ is a trivial vector bundle, it is enough to show that $\Lambda^2 T^* M$ is a trivial vector bundle. Now, we take local coordinates on $M$: let $(\theta, \phi)$ be parameters over the domain $(0, \pi) \times (-\pi, \pi)$ such that $$\begin{align} x & = \sin (\theta) \cos (\phi) & \mathrm{d} x & = \cos (\theta) \cos (\phi) \, \mathrm{d} \theta - \sin (\theta) \sin (\phi) \, \mathrm{d} \phi \\ y & = \sin (\theta) \sin (\phi) & \mathrm{d} y & = \cos (\theta) \sin (\phi) \, \mathrm{d} \theta + \sin (\theta) \cos (\phi) \, \mathrm{d} \phi \\ z & = \cos (\theta) & \mathrm{d} z & = - \sin (\theta) \, \mathrm{d} \theta \end{align}$$ One then computes that $$\frac{1}{z} \, \mathrm{d} x \wedge \mathrm{d} y = \sin (\theta) \, \mathrm{d} \theta \wedge \mathrm{d} \phi$$ but the RHS is manifestly non-zero everywhere in the domain of parametrisation. Repeating this calculation for the various permutations of $(x, y, z)$ and various other parametrisations then shows that $$\frac{1}{z} \, \mathrm{d} x \wedge \mathrm{d} y = \frac{1}{x} \, \mathrm{d} y \wedge \mathrm{d} z = \frac{1}{y} \, \mathrm{d} z \wedge \mathrm{d} x$$ whereever these expressions make sense.

Of course, one might ask where I conjured up such a $2$-form. The answer has to do with the Riemannian metric. In local coordinates, we have $$g = \mathrm{d} \theta \otimes \mathrm{d} \theta + (\sin \theta)^2 \, \mathrm{d} \phi \otimes \mathrm{d} \phi$$ and so we get a volume form $$\omega = \sin (\theta) \, \mathrm{d} \theta \wedge \mathrm{d} \phi$$ By general theory, this is known to be a local expression for a single nowhere-vanishing global section, so all that remains is to find expressions for it in other charts.


Now, some words about algebraic geometry. If $A$ is a nice ring – say a finitely-generated integral domain over a field $k$ – one can define the module of Kähler differential forms as follows: $$\Omega^1_{A / k} = \mathfrak{a} / \mathfrak{a}^2 \text{ where } \mathfrak{a} = \ker (f \otimes g \mapsto f g : A \otimes_k A \to A)$$ This is naturally an $A$-module, because $(A \otimes_k A) / \mathfrak{a} \cong A$. There is a canonical $k$-linear (not $A$-linear!) map $\mathrm{d} : A \to \Omega^1_{A / k}$ induced by the map $f \mapsto f \otimes 1 - 1 \otimes f : A \to A \otimes_k A$. One readily verifies that $\mathrm{d}$ satisfies the Leibniz rule, and it can be checked that $\Omega^1_{A/k}$ is generated as an $A$-module by $\{ \mathrm{d} f : f \in A \}$. So far so good. Then one can go on to define $\Omega^n_{A/k}$ to be the $n$-th exterior power (over $A$) of $\Omega^1_{A/k}$. It is a fact, at least when $k$ is algebraically closed and $A$ is a regular ring of dimension $n$, that $\Omega^n_{A/k}$ is a projective $A$-module of rank $1$.

Unfortunately, it is not at all clear from all this why $\Omega^2_{A / \mathbb{R}}$ should have a global generator of the form in the preceding discussion, when $A = \mathbb{R}[x, y, z] / (x^2 + y^2 + z^2 - 1)$. Yet it does, if we are prepared to regard $A$ as an affine scheme over $\operatorname{Spec} \mathbb{R}$: each of the three expressions is defined on a dense open subset of $\operatorname{Spec} A$, and they agree on the overlaps just as in differential geometry, so they patch together to give a global section that has non-zero stalks over all the points...

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Thank you so much for writing this all out. I was about to ask about $z=0$ when I first read your comment on the global generator, but I see that it's essentially just a patching argument. –  Andrew Parker Mar 19 '12 at 21:59
    
Though I'm a little confused about the generators of $\Omega^1_{A/k}$ that you mention. Certainly it's not d$f$ for all $f \in A$? More like {d$x_1$, d$x_2$,$\ldots$,d$x_n$} where $\{x_1,x_2,\ldots,x_n\}$ is a regular sequence in $A$? –  Andrew Parker Mar 19 '12 at 22:05
    
I didn't say they were linearly independent! –  Zhen Lin Mar 19 '12 at 22:13

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