# Tag Info

## New answers tagged exterior-algebra

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Clifford algebra might make this manipulation easier: its geometric product of vectors (or forms) is associative like the wedge product, but still has the metrical terms needed for the codifferential. In clifford notation, we have, $$\eta(\alpha) = X \rfloor (\nabla \rfloor \alpha) + \nabla \rfloor (X \rfloor \alpha)$$ You can see here that the ...

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The Moyal bracket is the calculational manifestation of noncommutative geometry in rough terms. Essentially, to do math over a noncommutative geometry one simply replaces regular products with $\star$ products which encode the deformation considered. However, much of the calculation goes through more or less the same. The Hodge $\star$ is completely ...

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Grassmann numbers can be embedded in a clifford algebra. Let $g(a,b)$ be some scalar-valued, symmetric, bilinear function for $a, b$ that are linear combinations of $\theta_i$--call $a,b$ "vectors". Then the multiplication laws could be tweaked to read $$\theta_i \theta_j + \theta_j \theta_i = g(\theta_i, \theta_j)$$ True grassmann numbers are the case ...

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There's a mistake in the last line in the calculation of $i_{\hat{X}_t}\beta$. The $t$'s in the numerators should be in the denominators: $$i_{\hat{X}_t}\beta = \frac{-zx}{\color{red}{t}r^3}\, dy + \frac{zy}{\color{red}{t}r^3}\, dx = \frac{z}{\color{red}{t}r^3}(y\, dx - x\, dy).$$ Then $$\Phi_t^*i_{\hat{X}_t}\beta = \frac{z}{(x^2 + y^2 + t^2 ... 2 The second definition is just wrong in characteristic 2. Any author who uses it is either implicitly ignoring the characteristic 2 case or is being imprecise. You find similar problems with the definition of Clifford algebras; there are again two variants and one of them is wrong in characteristic 2. 3 The first map is just the induced map on the 2-graded component of the exterior algebra on K^3, namely \Lambda^2(K^3). The second is the map on the 2 tensors in the tensor algebra, i.e. elements of K^3\otimes K^3. On the exterior product we have the diagonal action, so determine the action on the basis in which your matrix is written. If the ... 3 (i) Your understanding is correct. (ii) The significance of the ordering is mainly bookkeeping; in particular they provide an easy way to start from a basis e_1,\dots,e_n for the vector space V and extend it to a basis for the spaces of p-vectors and the exterior algebra. (iii) An algebra is a vector space with a product (that satisfies some rules). ... 1 If you're already using exterior algebra, clifford algebra isn't that much of a stretch to use. It just relies on an associative product called the geometric product. In an orthonormal basis, its properties are$$e_i e_j = \begin{cases} 1 & i = j \\ -e_j e_i & i \neq j \end{cases}$$Elements of the clifford algebra are members of a 2^n ... 0 Because the similarity between symmetric tensor product and wedge product, I will discuss only the wedge product here. It is common to see both two definitions of wedge product in different textbooks. Given two differential forms \alpha\in\bigwedge^p(V) and \beta\in\bigwedge^q(V), we can define the wedge product as \displaystyle ... 3 Since$$X\times Y=(X^2Y^3-X^3Y^2)e_1+(X^3Y^1-X^1Y^3)e_2+(X^1Y^2-X^2Y^1)e_3$$then$$(X\times Y)^i=\eta_{ijk}X^jY^k,$$is the correct formula for components. In the other hand if we agree the volume form be e^{*1}\wedge e^{*2}\wedge e^{*3} and defined by$$e^{*1}\wedge e^{*2}\wedge e^{*3}=\sum_{\sigma\in S_3}(-1)^{\sigma} e^{*\sigma(1)}\otimes ...

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A bilinear form on a vector space $V$ over a field $\Bbb{F}$ is a bilinear map $B:V\times V\to \Bbb{F}$ which eats a pair of vectors to give a scalar according to $(v,w)\mapsto B(v,w)$. This can be generalized.

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To elaborate on Zhen Lin's comment: either viewpoint of a bilinear form is okay due to the universal property of the tensor product. Given a vector space $V$ with field of scalars $k$, I've usually seen a bilinear form defined as a bilinear map $B : V \times V \to k$, so a function that eats two vectors as in your first definition. However, by the ...

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