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Questions tagged [exterior-algebra]

For questions on the exterior algebra, and related concepts such as the wedge product, the tensor algebra and differential forms.

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FEM vs DEC viewpoint [closed]

It occurred to me that the FEM (finite element method) and DEC (discrete exterior calculus) viewpoints are different, and I'm wondering if you agree with my comparison. Consider solving the Laplace ...
Zohar Levi's user avatar
4 votes
0 answers
43 views

Can we construct the exterior algebra just from simple multivectors?

$ \newcommand\K{\mathbb K} \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\Lip{\mathrm{Lip}} \newcommand\ev{\mathrm{ev}} \newcommand\Gr{\mathrm{Gr}} $Let $V$ be a finite-dimensional $\K$-...
Nicholas Todoroff's user avatar
2 votes
1 answer
35 views

Show that the dual Lefschetz operator applied to a two-form $\alpha$ is explicitly given by $\Lambda \alpha =\sum_i \alpha(x_i, y_i)$.

Choose an orthonormal basis $x_1, y_1 = J(x_1), \dots , x_n, y_n = J(x_n)$ of an euclidian vector space $V$ endowed with a compatible almost complex structure $J$. Show that the dual Lefschetz ...
Rene's user avatar
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Wedge product and isomorphism between $\bigwedge^{k}T_{p}^{*}M$ and $\left(\bigwedge^{k}T_{p}M\right)^{*}$

Wikipedia states that there is the following isomorphism, $$ \bigwedge^{k}T_{p}^{*}M \cong \left(\bigwedge^{k}T_{p}M\right)^{*} $$ More concretely, we think of the k-form, $\omega$ as either a linear ...
Jeff's user avatar
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Is the 'square matrix' in a 2d linear transformation just a bivector? [closed]

$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$ If so, how do we call the operation of ...
user66901's user avatar
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18 views

Exterior powers of Weyl algebra

I'm working with first Weyl algebra $\frac{k[x,y]}{<xy-yx-1>}$and I want to compute its cohomology using Koszul complex, so I need to find its exterior powers as a module over ground field k, ...
VadimStacheff's user avatar
3 votes
3 answers
157 views

What operation on matrices corresponds to the curl of a vector field?

Given the total derivative $Df$ of a (sufficiently) smooth function $f:\mathbb{R}^n \to \mathbb{R}^n$, the trace of the total derivative matrix corresponds to the divergence of $f$ (considered as a ...
hasManyStupidQuestions's user avatar
1 vote
0 answers
15 views

Question on homotopy for Koszul Complex

I am working on illustrating the relationship between the originally defined Koszul complex and its dual version by examining the definitions of the differentials we define for the Koszul complex. For ...
Beginner's user avatar
3 votes
1 answer
173 views

A simple question about the Hodge star

The usual definition of the Hodge star says that it maps $\Lambda^k(V)$ to $\Lambda^{n-k}(V)$ in such a way that for each pair $\omega, \eta \in \Lambda^k(V)$ holds $\omega \wedge *\eta = \langle \...
tsnao's user avatar
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Linear relations between minors of a matrix

Let $X = (x_{ij})_{1 \leq i, j \leq n}$ by an $n \times n$-matrix over the field $Q = \mathbf{F}(x_{ij})_{ij}$ for some field $\mathbf{F}$. Let $[n] = \{1,\dotsc,n\}$. For subsets $I, J \subseteq [n]$,...
Bubaya's user avatar
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Counting operator on $\Lambda^* V$.

Let $V$ be a finite-dimensional vector space. One defines the counting operator $H : \Lambda^* V \to \Lambda^* V$ by $H\vert_{\Lambda^{k}V}=(k-n)\cdot\text{id}$, where $\dim_\Bbb R V = 2n$. Anyone ...
Tim's user avatar
  • 187
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1 answer
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Dot product of two exterior products and associativity of geometric product?

This is a quick and basic question. I looked online (Wikipedia articles, Wolfram, etc..., and poked inside of Hestenes and Snygg's books, but couldn't easily pull out an answer). I'm going to define ...
Nate's user avatar
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2 votes
1 answer
113 views

Prove that $d_3d_3^*+d_3^*d_3=-\nabla^2$

Consider the geometry in $\mathbb R^3$, define $$d_3=dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}+dz\frac{\partial}{\partial z}.$$ We then define the Hodge star operator $*_3:\Omega^p(\...
Ho-Oh's user avatar
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2 votes
1 answer
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What's the definition of dual number at perspect of exterior algebra?

In Dual Number it said that "It may also be defined as the exterior algebra of a one-dimensional vector space with $\varepsilon$ as its basis element." But I can't find the detailed rigorous ...
Richard Mahler's user avatar
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Prove that the isotopy generated by a time-dependent symplectic vector field is a symplectomorphism

Let $M$ a compact and connected smooth manifold. Suppose $X_t$ is a time-dependent symplectic vector field and let $\phi_t$ be the isotopy generated by $X_t$. Prove that $\phi_t ∈ Symp(M, \omega)$ for ...
some_math_guy's user avatar
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24 views

How to determine that the quotient of the top power of two $2$-forms is $\leq 1$

For full context, I work on a complex manifold $M$ of dimension $n$ where I have a Hermitian $(1,1)$-form $\Omega$ and $(1,1)$-form $\tau$ which is negative semidefinite at a point. I am trying to see ...
rosecabbage's user avatar
  • 1,665
5 votes
1 answer
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Kernel of the action of GL(V) on exterior square of V

I wonder whether anyone knows a reference for the following result? I can give a shortish proof, but would prefer to cite the literature if possible. Theorem Let $V=F^n$ be an $n$-dimensional $F$-...
Glasby's user avatar
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1 answer
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Simplifying $d(\frac{2y_1}{r^2+1})\wedge d(\frac{2y_2}{r^2+1})\wedge\cdots\wedge d(\frac{2y_n}{r^2+1})$, where $r^2=\sum y_i^2$

I've been trying to find a nice, simple form for the following expression: $$d\left(\frac{2y_1}{r^2+1} \right)\wedge d\left(\frac{2y_2}{r^2+1} \right)\wedge ... \wedge d\left(\frac{2y_n}{r^2+1} \right)...
user13121312's user avatar
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52 views

Contraction of the determinant

In $\mathbb{R}^{n+1}$, we consider an orthonormal basis $e_0,\dots,e_n$ and $\alpha_0,\dots,\alpha_n$ its dual basis. Then, the determinant is $\alpha_0 \wedge \cdots \wedge \alpha_n$. I am seeking a ...
JulianDoyle's user avatar
0 votes
1 answer
69 views

Natural isomorphism between tensor product and exterior product

I am requesting help with the following problem. Below all rings are commutative with unit, and for a ring $R$, we define an $R$-algebra to be a ring $R'$ with ring homomorphism $f: R \to R'$. This ...
Abced Decba's user avatar
1 vote
1 answer
96 views

$\mathbb{C}-$linear extension of 2-forms to (1,1)-forms

I am trying to analyze a bit how we can extend a differential form to the complexification $V\otimes\mathbb{C}=V_{\mathbb{C}}$ of the vector space. Of course, you can do this for a general $k$-form, ...
領域展開's user avatar
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1 vote
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20 views

Two view points of exterior algebra - k-vectors and tangent spaces

I'm trying to see how two ideas are related: In (Discrete) Exterior Calculus, we defined k-vectors as volumes, and that k-forms are like "measurement tools", that allow us to measure k-...
blz's user avatar
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3 votes
1 answer
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L'Hopital's rule with dual numbers

Background: For the dual numbers, we extend the reals with an additional unit vector $\epsilon$ subject to the constraint that $\epsilon^2 = 0$. We can write dual numbers as $x_0 + x_1 \epsilon$ for $...
kc9jud's user avatar
  • 248
3 votes
1 answer
147 views

Hodge star operator evaluated at an orthonormal basis (Proposition 1.2.20 Huybrechts)

We want to show: $$*(e_{i_1} \wedge\ldots \wedge e_{i_k})=\epsilon e_{j_1} \wedge\ldots \wedge e_{j_{n-k}}$$ where $\epsilon$ is the sign of the permutation $i_1 \cdots i_k \bar i_1 \cdots \bar i_{n-k}...
領域展開's user avatar
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Fundamental form $\omega=\sum_{i\leq m}v^*_i\wedge (Jv_i)^* $ with a complex structure $J$

Let $V$ be a $\mathbb{C}-$ vector space, $J$ an almost complex structure on $V$ and take a real orthonormal basis $\langle v_1,Jv_1,\ldots,v_n,Jv_n\rangle $ with a scalar product $\langle,\rangle = \...
領域展開's user avatar
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0 votes
1 answer
60 views

Interior, cross and outer products between two multivectors?

For two arbitrary multivectors $\mathbf u$ and $\mathbf v$, what are the definitions of the interior (or scalar) product $\mathbf u\cdot \mathbf v$, the cross product $\mathbf u\times \mathbf v$ (if ...
HelloGoodbye's user avatar
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53 views

Exterior algebra decomposition natural projections $ Π^{p,q}:\bigwedge {V}^*_{\mathbb{C}} \to \bigwedge^{p,q}V$ (Huybrechts book)

Let $\bigwedge^{p,q}V:=\bigwedge^{p}V^{1,0}\wedge \bigwedge^{p}V^{0,1}$ and $\bigwedge {V}^*_{\mathbb{C}}=\bigoplus_k \bigwedge^{k} V_{\mathbb{C}} $. Then one defines the natural projection $$ Π^{p,q}...
領域展開's user avatar
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0 answers
39 views

A question on alternating product and $SL_n(R)$

I am reading Fulton and Harris representation theory. In the section 8.2 - Examples of Lie Algebras, while calculating the Lie algebra of $SL_n(\mathbb R),$ the author tell that by definition $$ A_t(...
Eloon_Mask_P's user avatar
2 votes
1 answer
33 views

Minors of change of basis matrix are Kronecker deltas

All the vector spaces I will consider in the following could be thought to have a base field of characteristic $ 0 $ (just to be safe). I know that if $ A\colon V\to W $ is a linear mapping between to ...
GeometriaDifferenziale's user avatar
5 votes
1 answer
197 views

Alternative of exterior power as a tensor algebra

This question is related to my previous question. The $n$-th exterior power of a vector space (over some field of characteristic zero) $U$ is a pair $(\wedge^{n}, \bigwedge^{n}U)$ where $\bigwedge^{n}...
Idontgetit's user avatar
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0 votes
1 answer
88 views

Does this alternating map satisfies universal property?

In what follows, $U$ is a complex vector space. I have the following abstract definition. Definition: Let $\wedge^{n}: \overbrace{U \times \cdots \times U}^{\text{$n$ times}} \to V$ be an alternating $...
Idontgetit's user avatar
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0 votes
0 answers
81 views

Wedge product of two functions

This is what I understand in the case of finite dimensions: Consider $1-$vectors in $R^2$. Given two $1-$vectors, $v_1$ and $v_2$, we can form a wedge product $v_1 \wedge v_2$, and if we are given $...
user2167741's user avatar
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0 answers
28 views

Does the Hessian correspond to the exterior derivative of the gradient 1-form? Or does its skew-symmetrization?

Question: Given a twice totally differentiable (not necessarily $C^2$) function $f: \mathbb{R}^m \to \mathbb{R}^n$, do its $n$ Hessian matrices correspond to the exterior derivatives of its $n$ ...
hasManyStupidQuestions's user avatar
2 votes
1 answer
73 views

Computing wedge product and exterior differential

I am studying for first time Smooth manifolds and I have some issues understunding the wedge product and the exterior differential since my teacher does not provide examples. For example, one of my ...
Daniel García's user avatar
1 vote
0 answers
91 views

Abstract and concrete wedge product

Some notation Let $ V $ be a finite dimensional vector space (say, over the real numbers). Let's suppose that all we know about the exterior power $ \bigwedge^k V^* $ of the dual space $ V^* $ of $ V $...
GeometriaDifferenziale's user avatar
0 votes
1 answer
64 views

How to define the determinant in a basis-independent form using alternate maps, without using properties of the „usual” determinant?

Let $V$ be a vector space of dimension $n$ over some field. Let $Alt_k(V)$ be the set of $k$-multilinear alternate maps. My question boils down to this, I think (although I will provide more context ...
rosecabbage's user avatar
  • 1,665
1 vote
1 answer
114 views

Riemannian geometry, manifolds and volume elements

I have two quesitons about a book by Nakahara: Geometry,topology and physics In the snippet below how do I compute that $$|\det(\frac{\partial x^\mu}{\partial y^\kappa}\frac{\partial x^\nu}{\partial y^...
user122424's user avatar
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1 vote
2 answers
95 views

What is the Grassmann algebra of $2\times 2$ complex matrices?

The traceless hermitian $2\times 2$ complex matrices form a real Euclidean space $\mathfrak E_3$ with dot product $a\cdot b:=\frac{1}{2}(ab+ba)/I$ where $I$ is the $2\times 2$ identity matrix. A ...
mma's user avatar
  • 2,033
1 vote
0 answers
57 views

Trying to understand two differential forms

I'm struggling with two particular differential forms: the $\flat$ 1-form and the $\natural$ 2-form defined as follows: Let $X = (X_1,X_2,X_3)$ be a vector field in an open subset $O$ of $\Bbb R^3$. ...
GreekCorpse's user avatar
0 votes
1 answer
71 views

Exterior powers and restriction of scalars

Let $ R $ and $ S $ be two commutative rings and let $ f\colon R\to S $ be a homomorphism of rings. For any $ S $-module $ N $ denote with $ f^*N $ the module obtained from $ N $ by restriction of ...
GeometriaDifferenziale's user avatar
1 vote
1 answer
46 views

Viewing vectors in a Clifford algebra as reflections

Let $Cl(s,t)$ be the Clifford algebra over $\mathbb{R}^{s,t}$ where $(s,t)$ is the signature of the bilinear form $\eta$. Let $Pin(s,t)$ be the associated pin group and define $$R: Pin(s,t) \times \...
CBBAM's user avatar
  • 6,019
1 vote
1 answer
76 views

Whys does Alt($\alpha)=\alpha \iff \alpha \in \Lambda^k (V^*) $hold for $ k=0 ?$

The following is according to Lee's introduction to smooth manifolds chapters 12 and 14. Let V be a vector space, $k\in \Bbb Z_{\ge 0}, \alpha $ a covariant k-tensor, i.e. $\alpha \in T^k(V^*)$ a ...
darkside's user avatar
  • 589
0 votes
1 answer
47 views

Computing $x\cdot e_n \cdot x \cdot e_n$ in a Clifford algebra over $\mathbb{C}^d$

I am closely following Hamilton's Mathematical Gauge Theory. Let $V$ be a vector space, $Q$ a bilinear form on $V$, and $CL(V,Q)$ the corresponding Clifford algebra. We can construct the Clifford ...
CBBAM's user avatar
  • 6,019
0 votes
0 answers
66 views

Contraction of a vector and an alternating tensor product of vectors

In Hamilton's Mathematical Gauge Theory he gives the following definition. Let $V$ be a finite-dimensional $\mathbb{K}$-vector space with a symmetric bilinear form $Q$. For $v \in V$ and $\sigma \in \...
CBBAM's user avatar
  • 6,019
0 votes
1 answer
47 views

interior product definition for $k>1$ $k$-vectors

On wikipedia exterior algebra page, section Hodge duality (https://en.wikipedia.org/wiki/Exterior_algebra#Hodge_duality) I read the following definition: ....Then the interior product induces a ...
LPSfr's user avatar
  • 1
1 vote
1 answer
54 views

How does the wedge product appear in the quotient approach to constructing an exterior algebra?

Let $V$ be a finite dimensional vector space and let $T(V)$ denote the tensor algebra $$T(V) = \bigoplus_{k=0}^\infty T^k(V)$$ where $$T^k(V) = \underbrace{V \otimes \cdots \otimes V}_k.$$ Let $I(V)$ ...
CBBAM's user avatar
  • 6,019
0 votes
1 answer
57 views

Why is the map $\text{Alt}^m V \times \text{Alt}^n V \to \text{Alt}^{m+n} V$ well defined?

In Fulton & Harris's "Representation Theory: A First Course" it is claimed that the wedge product $\wedge$ determines a product $\text{Alt}^m V \times \text{Alt}^n V \to \text{Alt}^{m+n} ...
רוי רז's user avatar
1 vote
2 answers
103 views

Rule for manipulating differential 2-forms in a physics derivation

This question comes from a more physics perspective, particularly thermodynamics and includes the usage of differential forms. We assume the existence of smooth functions with continuous derivatives ...
User198's user avatar
  • 168
0 votes
0 answers
79 views

Determinant Formula for Wedge Product via Universal Property of Exterior Powers

I'm currently learning about differential forms in my analysis class, and I thought I'd dig a bit more into the linear algebra of exterior powers. I've seen the universal property of $\bigwedge^k(V)$: ...
Joshua Yagupsky's user avatar
1 vote
0 answers
127 views

Cross-product Exercise

I am doing some problems in Spivak's calculus on manifolds, and I was stuck on this one. It says that: If $w_1, ... ,w_{n-1} \in \mathbb{R}^n$, show that $$| w_1 \times ... \times w_{n-1}| = \sqrt{\...
HtmlProg's user avatar

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