What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T00:32:14Z http://mathoverflow.net/feeds/question/95470 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95470/what-is-the-canonical-isomorphism-between-the-tensor-products-of-the-top-exterior What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces? Alberto Abbondandolo 2012-04-28T22:49:14Z 2012-04-29T01:51:09Z <p>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 $$\Lambda^{\max}(X) \otimes \Lambda^{\max}(Z) \cong \Lambda^{\max} (Y), \qquad x \otimes B_* (y) \rightarrow A_*(x) \wedge y$$</p> <p>or the isomorphism $$x \otimes B_* (y) \rightarrow y \wedge A_*(x).$$</p> <p>Here $x$ is a generator of $\Lambda^{\max}(X)$ and $y\in \Lambda^{\dim Z}(Y)$ is such that $A_*(x) \wedge y$ generates $\Lambda^{\max}(Y)$. </p> <p>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.</p> <p>Reasonable requirements that one should ask to this canonical isomorphism are:</p> <p>1) to the exact sequence $0\rightarrow X \stackrel{A}{\rightarrow} Y \rightarrow 0$ one associates the isomorphism $x \mapsto A_*(x)$;</p> <p>2) naturality with respect to isomorphisms of exact sequences.</p> <p>However, these requirements do not determine the isomorphism uniquely.</p> <p>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? </p> http://mathoverflow.net/questions/95470/what-is-the-canonical-isomorphism-between-the-tensor-products-of-the-top-exterior/95471#95471 Answer by Liviu Nicolaescu for What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces? Liviu Nicolaescu 2012-04-28T23:37:17Z 2012-04-29T01:51:09Z <p>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.</p> <p>Given a short exact sequence of finite vector spaces</p> <p>$$0 \to A\to B\to C\to 0$$</p> <p>we get a dual short exact sequence</p> <p>$$0\to C^* \to B^* \to A^* \to 0.$$</p> <p>For more details an a possible way out of this thorny situation see Section 1.2 of <a href="http://www.nd.edu/~lnicolae/Torsion.pdf" rel="nofollow">my book on torsion</a>. There I explain in some detail Deligne's rules of operating with determinant lines.</p> http://mathoverflow.net/questions/95470/what-is-the-canonical-isomorphism-between-the-tensor-products-of-the-top-exterior/95476#95476 Answer by Anton Fetisov for What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces? Anton Fetisov 2012-04-29T01:22:18Z 2012-04-29T01:22:18Z <p>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 <code>$$\Lambda^{*}{X} \otimes \Lambda^{*} {Z} \simeq \Lambda^{*} {Y}$$</code> Specializing to the subalgebra of topmost powers, we get the required canonical isomorphism.</p> <p>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 <code>$$(x \cdot y_1) \wedge y_2 = (y_1 \wedge x) \wedge y_2 = - (y_1 \wedge y_2 ) \wedge x = - x \cdot (y_1 \wedge y_2)$$</code></p>