Is there a preferable convention for defining the wedge product? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:41:18Z http://mathoverflow.net/feeds/question/54343 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product Is there a preferable convention for defining the wedge product? Giuseppe 2011-02-04T18:17:41Z 2013-02-22T14:38:34Z <p>There are different conventions for defininig the wedge product $\wedge$.</p> <p>In Kobayashi-Nomizu, there is $\alpha\wedge\beta:=Alt(\alpha\otimes\beta)$, in Spivak, we find $\alpha\wedge\beta:=\frac{k!l!}{(k+l)!}Alt(\alpha\otimes\beta)$, where $\alpha$ and $\beta$ are any forms of degree $k$ and $l$ respectively, and $Alt(\cdot)$ take the alternating part of the tensor.</p> <p>But, is there a rationale to prefer one of them among the others?</p> <p>If not, what do you prefer? and for what reason?</p> http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/54353#54353 Answer by Greg Kuperberg for Is there a preferable convention for defining the wedge product? Greg Kuperberg 2011-02-04T19:53:42Z 2011-02-04T19:53:42Z <p>The issue here is that there are really two tasks: (1) Define the algebra of differential forms on a manifold, and (2) implement them as multilinear functions on tangent vectors. The natural way to define them is along the lines of what Donu said: The differential forms at a point are like a polynomial algebra over the vector space <code>$V = T^*_pM$</code>, except supercommutative (or graded-commutative) instead of commutative. The supercommutativity condition is identical to taking a quotient of the tensor algebra, which is the free non-commutative algebra over the vector space $V$.</p> <p>But then for the second task, you would like a monomial, in a standard basis of cotangent vectors, to take values of ${0,1,-1}$ if you pair it with a standard basis of vectors. For example, you would like to say $$dx \wedge dy = dx \otimes dy - dy \otimes dx,$$ because that evaluates to $1$ on $(\hat{x},\hat{y})$ and $-1$ on $(\hat{y},\hat{x})$. In order to do this, you have to implement the wedge product with antisymmetrization and with factorials, actually the reciprocal of the factor you give: $$\alpha \wedge \beta = \frac{(a+b)!}{a!b!} \mathrm{Alt}(\alpha \otimes \beta).$$</p> <p>If I were explaining the subject, I would handle points (1) and (2) separately. It is common to conflate the two concerns. It amounts to either definition forms as a subspace of tensors (the usual solution), or as a quotient space of tensors. The real issue is that they need to be both, and that double role leads you to the factorial factors.</p> http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/54355#54355 Answer by Dan Lee for Is there a preferable convention for defining the wedge product? Dan Lee 2011-02-04T20:00:52Z 2011-02-04T20:00:52Z <p>It is a convention. It doesn't matter. This is like asking whether there should be a $2\pi$ or a $\sqrt{2\pi}$ in the definition of Fourier transform. In other words, not very interesting, and definitely not a research math question.</p> http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/54375#54375 Answer by MTS for Is there a preferable convention for defining the wedge product? MTS 2011-02-04T23:33:29Z 2012-02-16T20:15:26Z <p>I think a lot of people run into this issue. The way I think about it is the following:</p> <p>Take your finite-dimensional vector space $V$ and form its tensor algebra $T(V)$. Define $\mathcal{J}$ to be the 2-sided ideal in $T(V)$ generated by elements of the form $v \otimes v$, and then define the exterior algebra to be $\Lambda(V) = T(V) / \mathcal{J}$. This exhibits the exterior algebra as a quotient of the tensor algebra.</p> <p>The different conventions you see for the wedge product arise from different embeddings of the exterior algebra into the tensor algebra. Define on $V^{\otimes n}$ the map $$A_n (v_1 \otimes \dots \otimes v_n) = \frac{1}{n!} \sum_{\pi \in S_n} sgn(\pi) v_{\pi(1)} \otimes \dots \otimes v_{\pi(n)},$$ (or possibly with $\pi^{-1}$ instead of $\pi$, although I guess it doesn't matter) and then define on the tensor algebra the map $$A = \bigoplus_{n=0}^{\infty} A_n.$$ Then you can show easily that $A_n^2 = A_n$ for all $n$, so that $A$ is a projection.</p> <p>The point is that $\mathcal{J} = \mathrm{ker} (A)$, so that you can identify the quotient $\Lambda(V)$ with $\mathrm{im} A$, i.e. we have now embedded the exterior algebra as a subspace of the tensor algebra. This is where the two conventions differ. I have defined $A_n$ with a $\frac{1}{n!}$ in front, but some don't do so. Of course, this doesn't change the kernel of the map, but it does change the embedding of the exterior algebra into the tensor algebra.</p> <p>The important point is that $A$ is not an algebra map of $T(V)$ to itself, so the embedding $\Lambda(V) \to T(V)$ is not an embedding of algebras. Now you ask how to describe the exterior product in terms of the product in $T(V)$. Take $\alpha \in \Lambda^k(V)$ and $\beta \in \Lambda^l(V)$ with representatives $\tilde{\alpha} \in \mathrm{im}(A_k)$ and $\tilde{\beta} \in \mathrm{im}(A_l)$, respectively. Then $A_{k+l}(\tilde{\alpha} \otimes \tilde{\beta})$ is the representative of $\alpha \wedge \beta$ that you're looking for.</p> <p>Essentially, it boils down to whether or not you put the $\frac{1}{n!}$ in front of your alternating map or not.</p> http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/54407#54407 Answer by Donu Arapura for Is there a preferable convention for defining the wedge product? Donu Arapura 2011-02-05T15:06:45Z 2011-02-06T12:53:04Z <p>The answers by Greg and MTS are quite thorough, so there is not much more to say about that. However, I would like to explain my comment that viewing differential forms as antisymmetric tensors is often inadvisable, although I don't want to seem too dogmatic about this.</p> <p>My first argument is pedagogical. Making the above identification can be confusing (as evidenced by the question) and is frequently besides the point. A few years back, when I taught a vector calculus class, I decided to do differential forms. The students had no idea about tensor products or multilinear algebra, so it would have been a bad idea to attempt this approach. Instead, I told them that $dx$ etc. were symbols subject to the chain rule $dx = \frac{\partial x}{\partial u}du+\ldots$, and that they could be multiplied in such a way that $dx\wedge dy= - dy\wedge dx$. I gave a heuristic explanation in terms of oriented areas of "infinitesimal" rectangles for why this should be so... I won't claim that the experiment was entirely successful, but it could have been a lot worse.</p> <p>My second argument is more mathematical. Differential forms can be defined within algebraic geometry for quite general spaces. Here the approach using antisymmetric tensors can lead to serious problems. In characteristic $p>0$, the denominators will be undefined in general. As an interesting side note, in the algebraic proof of the Hodge theorem by Deligne and Illusie they do find it necessary to make this identification. But they have to restrict the dimension of the space to be less than $p$ for precisely this reason. Although in the limiting case of characteristic $0$, this is a nonissue.</p> http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/54458#54458 Answer by Qfwfq for Is there a preferable convention for defining the wedge product? Qfwfq 2011-02-05T19:40:38Z 2011-02-05T19:40:38Z <p>Each one of the two conventions has it's own advantage: the one with the normalizing coefficient makes the exterior algebra sit inside the tensor algebra (as the subspace of alternating tensors) and the "Alt" map be a projection onto that subspace hence the identity on alternating tensors, while the convention with*out* the normalizing factor is better suited for a ground field of positive characteristic as otherwise the denominator of the normalizing factor would be zero.</p> http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/122631#122631 Answer by Martin Gisser for Is there a preferable convention for defining the wedge product? Martin Gisser 2013-02-22T14:38:34Z 2013-02-22T14:38:34Z <p>I prefer (alas!) the Kobayashi-Nomizu "algebraic" version. Besides the <em>formal benefits</em> of having a projector and not needing to carry around combinatorial factors, here is a <em>differential geometric case</em>:</p> <p>If $\nabla \alpha$ is the Levi-Civita connection applied to the 1-form $\alpha$, then the exterior differential $d\alpha$ is the antisymmetric part of $\nabla\alpha$. In the Spivak "geometer" version it would be $\frac 1 2 d\alpha$.</p> <p>(The symmetric part of this decompostion involves the Lie derivative of the metric. See the nice book by W.A. Poor, Differential Geometric Structures, or Peter Petersen's Riemannian Geometry 2nd ed. (Who has found this wondrous decomposition?))</p> <p><strong>I hope I'm confused on this...</strong> 1) Two different "canonical" exterior differentials would be quite a scandalous mess. 2) I don't know any textbook mentioning this issue</p> <p>P.S.: see appendix of <a href="http://arxiv.org/abs/math-ph/0212043" rel="nofollow">http://arxiv.org/abs/math-ph/0212043</a> showing what bad can happen</p>