# Tagged Questions

It is a quotient - of the tensor algebra, obtained by taking graded sum over whole numbers $n$ of $n$-fold tensor products - by the ideal generated by elements of the form $a\otimes a$. We write the residue class of $a\otimes b$ in this algebra, as $a\wedge b$ and call it the wedge product.

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### Exterior product in 3 dimension

I'm curious about the exterior product in $3$ dimensions. Is it the same as the cross product? If it is, does anybody here have an idea about its calculation? $(a \wedge b) \wedge c = ?$
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### wedge product of $m$ vectors in $\mathbb{R}^n$

I came across the symbol $|v_1 \wedge \dots \wedge v_m|^{-1}$ in a paper - this is the norm of the wedge product of vectors $v_k \in \mathbb{R}^n$ . I thought it's meaning was self-evident until I ...
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### Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?

Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$? I know that ...
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### Geometric introduction to exterior algebra

Could anyone point me to a geometric introduction to exterior algebra (meaning, one with a good number of figures and/or verbal descriptions of geometric objects in it)? Thanks!
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### Norm inequality with wedge product

Anyone could help me to prove this following inequality? $\displaystyle\frac{||(u+v)\wedge w||}{||u+v||}\le \frac{||u\wedge w||}{||u||} +\frac{||v\wedge w||}{||v||}$ where $u\wedge v$ is the wedge ...
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### Tensor products, existence of a unique linear map

Question: Given a bilinear map $B: V\times W\to X$, show there exists a unique linear map $T:V \otimes W\to X$ s.t. $B= T \circ \phi$ Background: We define $V \otimes W$ by F[ ...
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### How to compute Ext over an exterior algebra

I found this question in several places (even on mathoverflow and mathstackexchange), but I never found a satisfying answer. Let $k$ be a field and $V$ a finite dimensional $k$-vectorspace. I ...
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### Exterior power and alternating forms: explicit computations

I would like to get a more concrete understanding of a general isomorphism I have read about. I apologize if this is too basic, but I was not satisfied with the references at my disposal. Let $K$ be ...
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### Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

What does an exterior (second) derivative such as in $d^2=0$ have to do with second derivatives as in single- or multi-variable calculus? Is this a correct start: Calculus derivatives are good for ...
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### Wedge product of 0-form with 1-form

What is the wedge product $\wedge$ of a $0$-form $f(x_1,...,x_n)$ with a $1$-form $\displaystyle\sum_{i=1}^{n} a_i dx_i$? According to the theory, it should be a (0+1=1)-form.
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### Exterior product of $\Bbb Z[x,y]$

Let $R$ be the polynomial ring $\Bbb Z[x,y]$ in the variables $x,y$. If $M=R$ (so we are considering the $R$-module $R$) then $\wedge^2 (M)=0$. This was an example in Dummit and Foote's Abstract ...
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Basic setting: Let $V$ be a $k$-vector space of finite dimension and $V^*$ its dual space. Let $\bigwedge^n V$ denote the $n$-th exterior power of $V$. Now the canonical pairing $$V \times V^{*} ... 1answer 343 views ### Effect of pullback of differential forms on an ideal Say that the exterior differential system (EDS) corresponding to a PDE system is:$$df-f_x\,dx-f_y\,dy-f_w\,dw-f_z\,dz=0,\\ a_1\,f_x+a_2\,f_y=0,\tag{sys}$$Of course we also require the independence ... 1answer 26 views ### To show that \Lambda^pL(V\rightarrow W) and L(\Lambda^pV\rightarrow W) are not necessarily isomorphic Let V and W be two vector spaces. Use L(V\rightarrow W) to represent the vector space of linear map from V to W. It is proved that \Lambda^p(V^*)\cong (\Lambda^pV)^*, where \Lambda is ... 0answers 403 views ### Hodge-Star-Operator on arbitrary oriented basis Assume that V is oriented finite dimensional vectorspace with dimension n, g \in T^0_2(V) a given symmetric and nondegenerate tensor. Let \mu be the corresponding volume element of V. ... 0answers 112 views ### Kernel of the Lie bracket [,]\colon\wedge^2\mathfrak g\to\mathfrak g I believe the following is probably well-known, but so far I couldn't find the answer by myself: Let \mathfrak g be a real (finite-dimensional) Lie algebra, and \wedge^2\mathfrak g its second ... 2answers 183 views ### Universal Property of the Exterior Algebra Let k be a field and let A be a commutative algebra over k. I want to calculate the exterior algebra \Lambda_A^\bullet A. We have \Lambda_A^0 A = \Lambda_A^1 A= A, and \Lambda_A^k A = 0 ... 2answers 101 views ### How to integrate this differential form on the boundary of the cube The setup. Assume u = u_1+iu_2: \mathbb{R}^3 \to \mathbb{C} and we have the differential 1-forms$$ \star\xi=-x_2 dx_3 + x_3 dx_2 $$and$$ u \times du = \sum_{i=1}^3 (u \times \partial_i u) dx_i = ...
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Let $V$ be a graded vector space, thought of as a collection $\{ V^n \}_{n \ge 0}$ of vector spaces. Let $V_{odd} = \bigoplus_{n \text{ odd}} V^n$ and $V_{even} = \bigoplus_{n \text{ even}} V^n$. I am ...
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### Transformation rule for a wedge product

Suppose two sets of covectors on a vector space $V, \beta^1,...\beta^k$ and $\gamma^1,...,\gamma^k,$ are related by $$\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$$ where $i=1,...,k$, for a $k\times k$ ...
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### Pullback expanded form.

Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$ According to Daniel ...
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### Lie Derivative for Wedge Product of Vector Fields

I am having trouble here. The context is: Let $X$, $Y$ and $S$ be vector fields ina a manifold (we can assume it's $\mathbb{C}^2$ though I'm pretty sure this should work in any manifold), and we can ...