Tag Info

Hot answers tagged

9

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this). If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called ...


4

Let $T(v_i) = \sum_j a_{ji} \cdot v_j$, so that $(a_{ji})$ is the matrix of $T$ w.r.t. to $(v_1,\dotsc,v_n)$. The wedges $v_{i_1} \wedge \dotsc \wedge v_{i_k}$ with $i_1<\dotsc<i_k$ form a basis of $\Lambda^k(V)$. We have: $$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} ...


3

$E^{\otimes 0}$ should be the identity of $\otimes$, i.e. the base ring $R$.


3

As you suspect the answer is yes, and you basically explain why it is so. If you are not convinced you could add the following details: your PDE $a_1 f_x + a_2 f_y=0$ is first order with the variable $f$ the dependent variable while $x,y,w,z$ are the independent variables. Standard coordinates in first jet space are $x,y,w,z,f,f_x,f_y,f_w,f_z$ while the ...


3

$$(dx \wedge dy)(\frac{\partial}{\partial(x)}) = (-dy \wedge dx)(\frac{\partial}{\partial(x)}) = -dy$$ since $$dx_j(\frac{\partial}{\partial(x_i)}) = \delta_{i,j}\ \ \text{ and }\ \ \ \ \ \ dx \wedge dy = - dy \wedge dx$$.


3

We have $f(1)^2=0$, hence by $A$-linearity $f(b) f(d) = bd f(1)^2 = 0$. By the way, the exterior algebra of $A$ over $A$ is $A[x]/(x^2)$, the algebra of dual numbers, with $x$ in degree $1$.


3

Here's a hint: For any vectors $\mathbf v,\mathbf w\in\Bbb R^3$, $$\omega(\mathbf x)(\mathbf v,\mathbf w) = \det(\mathbf x,\mathbf v,\mathbf w).$$ Here I'm writing $\mathbf x=(x,y,z)$. ADDENDUM: Here's a suggestion to address your uniqueness question. Consider two types of matrices with determinant $1$: $$\begin{bmatrix} c^2 & 0 & 0 \\ 0 & 1/c ...


3

It's easy to see that $\wedge^n R\cong \bigotimes^nR/I$, where $I=\{r_1\otimes...\otimes r_n|r_i=r_j $for $ $some $ i\neq j $}. Now $\bigotimes^2 R=(r_1\otimes r_2|r_i\in R)=(r_1r_21\otimes1|r_i\in R)\subset I$. This means that $\wedge^2 R\cong \bigotimes^2R/I=0$


3

It is an isomorphism, where $A$ is a commutative ring, and $V$ and $W$ are $A$-modules. I think we need $V$ and $W$ to be finitely generated and projective (but I'm not sure about this; perhaps someone can opine conerning this). See Theorem 7 here in Bergman's notes for more details. His proof includes a description of the inverse map. (In his notes, $k$ ...


3

Let $A$ be a commutative ring. Then $\Lambda(-)$ is a functor from $A$-modules to graded-commutative $A$-algebras which is left adjoint to the functor which takes the degree $1$ part. Because it is left adjoint, it preserves colimits, in particular coproducts. It follows $\Lambda(M \oplus N) \cong \Lambda(M) \otimes \Lambda(N)$. Looking at the $n$th degree ...


2

A $0$-form is nothing but a function. The exterior product is in this case the usual multiplication, i.e. the product is $$f(x)\wedge\sum_{i=1}^n a_i(x) dx_i=\sum_{i=1}^n a_i(x)f(x) dx_i$$ As you can see, this really is a $1$-form.


2

Just compute the dimensions. Suppose $\dim V = n$ and $\dim W = m$. On the one hand, $\Lambda^p L (V \to W)$ has dimension $\frac{(n m)!}{p! (n m - p)!}$; on the other hand, $L (\Lambda^p V \to W)$ has dimension $\frac{n! m}{p! (n - p)!}$. Of course, these two quantities are equal when $m = 1$, but already for $m = 2$ we have $\frac{(2 n)!}{p! (2 n - p)!}$ ...


2

Good question! Here's a start. The ordinary derivative in one-variable calculus is a Lie derivative along a special vector field on $\mathbb{R}$; in particular, it is not a special case of the exterior derivative. The exterior derivative is instead some kind of "universal derivative": it records all of the information you would need to determine the ...


2

As remarked by Daniel Rust the area of the hexagon in question is the sum of the projected areas of the three "kinds" of cube facets. When a piece of a plane $\Sigma$ is orthogonally projected onto another plane $\Pi$ then the area is multiplied by $|\cos\phi|$, where $\phi$ is the angle between the planes, or equivalently: the angle between the ...


2

I think you might just be forgetting that $f$ is $A$-linear. That makes $af(d)=f(ad)$ and $f(b)c=f(bc)$ and $f(b)f(d)=bdf(1)^2=0$. An easy visualization of the operation on $A\oplus A$ that you described is as the subring of the $2\times2$ upper triangular matrices over $A$ with the elements $$ \left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\mid ...


2

being encouraged by Georges Elencwajg :) the simplest example I can come up with is $E$ with basis $a,b,c,d$ and the left ideal generated by $a+bc$; one easily checks that $(a+bc)d\notin I$. How to find it: $\bigwedge E$ is noncommutative, but it is graded-commutative. Ideals generated by $\mathbb Z/2\mathbb Z$-homogeneous elements are thus going to be ...


2

This is false without compactness. Every $2$-form on $\Bbb R^2-\{0\}$ is exact, but the infamous $1$-form $d\theta$ is not.


2

The norm is the same as the absolute value of the cross product followed by the dot product (or the absolute value of the determinant of a matrix with columns $a,b, c$): $$\| a \wedge b \wedge c \| = \lvert a \cdot (b \times c) \rvert = \lvert\det[ a\ b\ c]\rvert$$ Its geometric interpretation is quite different however. That is $a \wedge b \wedge c$ is ...


1

First, let's settle the issue of the empty tensor product. Consider the identity map $id: W\to W$ for an arbitrary $R$-module $W$. This is an $R$-balanced map out of the collection $\{W\}\cup \emptyset$ and so descends to a map $W\otimes E\to W$. But we know that $(W,id)$ satisfies the universal property of the tensor product of $W$, so $W\cong W\otimes E$ ...


1

Thanks to Sanchez I got the answer: First of all, we prove that the pairing on the exterior power of $V$ is invariant under the $G$-action: \begin{align*} P(gv,gv^*) & = P(gv_1 \wedge \dots \wedge gv_n, gv_1^* \wedge \dots \wedge gv_n^*) \\ & = \det ((g \cdot v_i^*)(g \cdot v_j))_{i,j} \\ & = \det (v_i^*(g^{-1} g \cdot v_j))_{i,j} \\ & ...


1

Edit: (Short version over Noetherian Domain) Exactly when M is projective One "extension" of your statement is that if $M$ has rank $r$, then $\mbox{rank}\wedge^k M = {r \choose k}$. In particular, when $k > r$ the module $\wedge^k M$ has rank $0$. This holds in any commutative ring for any finite rank module. (It comes considering $R^r \to M$, and ...


1

I thought it might be worthwhile to add an explicit example. Consider $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ with $$ [T] = \left[ \begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]$$ In particular, using the usual $e_1 = [ 1,0,0 ]^T=(1,0,0)$, $e_2 = (0,1,0)$, $e_3=(0,0,1)$ we have $$ T(e_1) = (a,b,c), \ \ \ ...


1

A toy example. Given an orthonormal basis $e_1,\dots,e_n$ for the vector space $V$, then $\nu=e_1\wedge\dots\wedge e_n$ and $$*(e_1\wedge e_2)=e_3\wedge\dots\wedge e_n\in \bigwedge^{n-2}V, $$ with $*e_1=e_2\wedge\dots\wedge e_n$ and $*e_2=-e_1\wedge e_3\wedge\dots\wedge e_n$ both in $\bigwedge^{n-1}V$. Introducing the insertion operator $i_\bullet: ...


1

I think the solution is not possible, though with you may have a solution to similar systems: By 1), $A$ must be a function, i.e., a $0$-form, and , since the wedge of functions is their product, you must have $A==\pm 1$, But then, substituting in 2), you will have $\pm\eta+\pm\eta=\eta$, which is not possible.


1

Let $\mathbf{n}$ be a unit vector in $\mathbb{R}^3$ be normal to the plane $P$. The projection $p$ of a vector $v\in\mathbb{R}^3$ onto the plane $P$ is given by $p(v)=v-(v\cdot\mathbf{n})\mathbf{n}$. The projection of the unit cube onto $P$ along the vector $\mathbf{n}$ is a hexagon whose interior has preimage which intersects the boundary of the unit cube ...


1

Let us consider the following general setting. We need it to prove the statement. The dg Lie algebra of poly-vector fields Let $M$ be a real manifold of dimension $n$ over the ground field $\mathbb K$. Let $$\operatorname{T}^{\bullet}_{poly}(M):=\mathcal C^{\infty}(M)\otimes_{\mathbb K}\wedge^{\bullet+1}\operatorname{T}(M) $$ be the algebra of poly ...


1

That $d^2 = 0$ is probably something you were already taught in vector calculus. For instance, you probably remember that $\nabla \times \nabla \phi = 0$ for $\phi$ a scalar field, or that $\nabla \cdot (\nabla \times E) = 0$ for $E$ a vector field. It's a good exercise to show that both of these can be written as $d^2 f = 0$ and $d^2 E = 0$. Of course, ...


1

The exterior derivative is a sort of infinitesimal operation. See for example Bachman's notes on differential forms. (See also comments to the question, above).


1

For any $0$-form such as $f(x_1, x_2, . . . , x_n)$ and any $k$-form $\beta$, the operation $f \wedge \beta$ of wedge multiplication of $\beta$ by $f$ is simply ordinary multiplication by $f$: $f \wedge \beta = f\beta$, a $k$-form. Thus $f \wedge \sum_1^n a_i dx_i = \sum_1 ^n fa_i dx_i, \tag{1}$ clearly another $1$-form. Details may be found at ...



Only top voted, non community-wiki answers of a minimum length are eligible