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One often reads (and writes) that an exact sequence of finite dimensional vector spaces $$ 0 \rightarrow X_1 \rightarrow X_2 \rightarrow \dots \rightarrow X_n \rightarrow 0 $$ induces a canonical isomorphism $$ \bigotimes_{i \; \mathrm{odd}} \Lambda^{\max} (X_ i) \cong \bigotimes_{i \; \mathrm{even}} \Lambda^{\max} (X_i), $$ where $\Lambda^{\max}(X)$ denotes the top exterior power of the vector field $X$. My problem is that there seem to be too many choices for the sign of this ``canonical'' isomorphism. For instance, to the exact sequence $$ 0 \rightarrow X \stackrel{A}{\rightarrow} Y \stackrel{B}{\rightarrow} Z \rightarrow 0 $$ it seems equally canonical to associate the isomorphism $$ \Lambda^{\max}(X) \otimes \Lambda^{\max}(Z) \cong \Lambda^{\max} (Y), \qquad x \otimes B_* (y) \rightarrow A_*(x) \wedge y $$

or the isomorphism $$ x \otimes B_* (y) \rightarrow y \wedge A_*(x). $$

Here $x$ is a generator of $\Lambda^{\max}(X)$ and $y\in \Lambda^{\dim Z}(Y)$ is such that $A_*(x) \wedge y$ generates $\Lambda^{\max}(Y)$.

Since this canonical isomorphism is often used in the theory of determinant bundles in order to define orientations for geometric objects, I find this uncertainty on a sign disturbing.

Reasonable requirements that one should ask to this canonical isomorphism are:

1) to the exact sequence $0\rightarrow X \stackrel{A}{\rightarrow} Y \rightarrow 0$ one associates the isomorphism $x \mapsto A_*(x)$;

2) naturality with respect to isomorphisms of exact sequences.

However, these requirements do not determine the isomorphism uniquely.

My question is: is there a standard convention regarding the definition of the canonical isomorphism which is associated to exact sequences of arbitrary length? And if not, what would be reasonable requirements to add to 1) and 2) in order to have a good definition?

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Probably a silly question, but why is in your example the second morphism 'equally canonical' in comparison with the first? In the second choice of isom. you do reverse the ordering between the factors, while the first isom. keeps it as it was. –  René Apr 28 '12 at 23:09
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To René: you are right, the first definition looks more "canonical". And this definition extends to sequences of arbitrary length. But is it the standard one? And how is it characterized? My reason to doubt that this is the correct definition comes from the construction of the determinant bundle on the space of Fredholm operators: in this construction one uses an exact sequence with 4 terms and the isomorphism which one associates to it is not this one (see D. QUILLEN, Determinants of Cauchy-Riemann operators over a Riemann surface, Functional Anal. Appl. 19 (1985), 31–34). –  Alberto Abbondandolo Apr 28 '12 at 23:30
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2 Answers 2

The isomorphism you stated exist not only on the level of topmost exterior power, but also on the level of the whole exterior algebra, considered as a superalgebra. For an exact sequence $$ 0 \to X \xrightarrow{A} Y \xrightarrow{B} Z\to 0 $$ it is defined in a similar way: $$ x\otimes B(y) \mapsto A(x) \wedge y $$ but here $x$ and $y$ are elements of $X$ and $Y$ respectively. That is, they are the generators of superalgebras $\Lambda^* {X}$ and $\Lambda^* {Y}$. Now this isomorphism is canonical, if you fix a certain definition of tensor algebra. A common way to define multiplication in tensor product of superalgebras is $$ (a\otimes b)(c\otimes d) = (-1)^{\deg b \deg c}(ac)\otimes (bd)$$ Here $a,b,c,d$ are some homogeneous elements of respective algebras. It is clear that a natural isomorphism must map generators to generators. Also, $$ a\otimes b = (a\otimes 1)(1\otimes b) $$ Mapping multiplication to multiplication and $x\otimes 1 \mapsto A(x)\in \Lambda^* (Y)$, $1\otimes B(y) \mapsto y\in \Lambda^* (Y)$, we get a uniquely defined isomorphism $$ \Lambda^{*}{X} \otimes \Lambda^{*} {Z} \simeq \Lambda^{*} {Y}$$ Specializing to the subalgebra of topmost powers, we get the required canonical isomorphism.

The reverse isomorphism you mentioned would be an isomorphism with some algebra $\Lambda^\prime Y$, that has the same multiplication for elements in $X$, but $x \cdot y = y \wedge x$ for $y \perp x$. This algebra isn't associative, because $$ (x \cdot y_1) \wedge y_2 = (y_1 \wedge x) \wedge y_2 = - (y_1 \wedge y_2 ) \wedge x = - x \cdot (y_1 \wedge y_2) $$

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I don't think that there is a "canonical" choice, and I will point to another headache. Is there a " natural isomorphism $\det X\cong \det X^*$? If so, how does this work in the following situation.

Given a short exact sequence of finite vector spaces

$$ 0 \to A\to B\to C\to 0 $$

we get a dual short exact sequence

$$ 0\to C^* \to B^* \to A^* \to 0. $$

For more details an a possible way out of this thorny situation see Section 1.2 of my book on torsion. There I explain in some detail Deligne's rules of operating with determinant lines.

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Thanks for the link to your book. If I understand well, the notion of "weighted line" allows $\Lambda^{\max}(X)$ to "remember" the dimension of the space $X$ and this helps when dealing with sign issues. –  Alberto Abbondandolo Apr 29 '12 at 7:47
    
Yes, and unfortunately, this cannot be helped. Think of doing intersection of two odd dimensional cycles of complementary dimensions in an even dimensional manifold. –  Liviu Nicolaescu Apr 29 '12 at 9:51
    
If X is a line, then det(X)=X and det(X*)=X*, but there is no canonical isomorphism between a line and its dual (e.g., a line bundle over a holomorphic manifold need not be isomorphic to its dual). A more natural statement would be (det(X))*=det(X*). The canonical pairing between det(X) and det(X*) can be defined using the canonical pairing between ΛX and Λ(X*). –  Dmitri Pavlov Apr 29 '12 at 12:37
    
@Dmitry That is true, but in applications one is given either a basis of $X$ (this happens when one computes torsions of based acyclic complexes) or an orientation of $X$, i.e., a basis of $X$ defined up to a positive multiple. Then you run into a bit of trouble with the conventions. –  Liviu Nicolaescu Apr 29 '12 at 14:03
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