Tag Info

Hot answers tagged

4

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 ...


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

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


3

I'll write out an answer for a form of three variables. This might yield some intuition. Thing to notice is that $dx^i\wedge dx^i=0$, and that $dx^i\wedge dx^j=-dx^j\wedge dx^i$, and $f_{xy}=f_{yx}$. Thus, because $dx^i\wedge dx^i=0$ $$d^2f=f_{xy}dy\wedge dx+f_{xz}dz\wedge dx+f_{yx}dx\wedge dy+f_{yz}dz\wedge dy+f_{zx}dx\wedge dz+f_{zy}dy\wedge dz $$ ...


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

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

At a non-critical point $x$ of $f$, the kernel of the $1$-form $\text{d}f$ is the tangent space to the surface $X := f^{-1}(f(x))$ at $x$; similarly, for a non-critical point $x$ of $g$, $\text{d}g_x$ has the tangent space of $Y := g^{-1}(g(x))$ at $x$ as its kernel. Putting both together, at non-critical points $x$ for both $f$ and $g$ the radical of the ...


3

The statement in the article has to be taken with a huge grain of salt. Even for $1$-forms, there's no consistent way to interpret all wedge products of $1$-forms as intersections. If you restrict attention to nonvanishing decomposable forms (ones that can be written locally as wedge products of $1$-forms), and add a few extra restrictions, then there is ...


2

Define the characteristic vector (fields) of a form as $\iota_v\omega=0$. By the properties of the interior product if both $v_1$ and $v_2$ are c.v., then so are their linear combinations, $\text{span}(v_1,v_2)$. Furthermore, if the rank of the form $r$, the number of linearly independent characteristic vectors is $n-r$. One can describe a tangent plane to ...


2

Let $X=z(x,y,z)/r^2$. Then $\beta=i(X)\Omega$ where $\Omega=dx\wedge dy \wedge dz$ is the volume element of $\mathbb{R}^3$ and $i$ is the interior product. Then the definition of divergence operator says $$d(i(X)\Omega)=\operatorname{div}(X)\Omega$$ So to show $d\beta=0$ is equivalent to show $\operatorname{div}(X)=0$ and we have ...


2

Perhaps a more understandable way to see this is to parametrize $\Bbb R^3-\{0\}$ by spherical coordinates: $$(x,y,z) = f(r,\phi,\theta)= (r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi).$$ Then $f^*\beta = (r\cos\phi)\dfrac{r^3\sin\phi\, d\phi\wedge d\theta}{r^4} = \cos\phi\sin\phi\,d\phi\wedge d\theta$. So it's clear, then, that $f^*(d\beta) = ...


2

Note that, in each term, we've written for instance $df \wedge dy \wedge dz$. Because $df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z} dz$, and $dy \wedge dy = 0, dz \wedge dz = 0$, we see that $$df \wedge dy \wedge dz = \frac{\partial f}{\partial x}dx \wedge dy \wedge dz.$$ So ultimately for this ...


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 ...


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

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

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

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

On the Visualisation of Differential Forms A Geometric Approach to Differential Forms Teaching Electromagnetic Field Theory Using Differential Forms


2

Incomplete thoughts: Thinking of the elements as written in an $n \times m$ grid, send it to the element $\alpha = \bigotimes_{j=1}^m (v_{1j} \wedge \cdots \wedge v_{nj}) \otimes \bigotimes_{i=1}^n (w_{i1} \wedge \cdots \wedge w_{im})$. (so, putting the $v$'s together in columns and the $w$'s in rows.) To see that this is well-defined, it suffices to ...


2

Hint: Write all vectors in the following form: $\pmatrix{1\\2\\3}=e_1+2e_2+3e_3$. For the tensor product, keep in mind that $e_1\otimes e_2$ is a linearly independent element to $e_2\otimes e_1$ (so they are as much different as they can), while for the exterior product we have $$e_1\land e_2\ +\ e_2\land e_1\ =\ 0\,.$$ That's all we have to know for this ...


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

If $\omega$ and $\tau$ are closed forms (which you have not used), then $$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$ hence $\omega \wedge \tau$ is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of $\omega$ and $\tau$. Let us suppose that $\omega = ...


2

Now apply the magic formula again: $$ L_X(\eta)=i_Xd\eta+di_X\eta $$ where $\eta=i_Yd\omega+di_Y\omega$ and get $$ L_X(L_Y\omega)=i_Xd(i_Yd\omega+di_Y\omega)+di_X(i_Yd\omega+di_Y\omega) $$ use that $d$ is linear $$ i_Xd(i_Yd\omega)+i_X(di_Y\omega)+di_X(i_Yd\omega)+di_X(di_Y\omega) $$ Subtract $L_Y L_X\omega$ and compare with $$ ...


2

In general, let $V$ be a real or complex vector bundle of dimension $n$. What can we say about $V$ if it admits $k$ independent nonvanishing global sections? This is equivalent to admitting a splitting $V \cong W \oplus \mathbb{R}^k$ resp. $V \cong W \oplus \mathbb{C}^k$, and so it implies some conditions on characteristic classes. If $V$ is real, then ...


1

More generally, if $f : M \to N$ is a homomorphism between free modules of rank $n$, then we can define its "adjugate homomorphism" $\overline{f} : N \to M$ to be the composition $$N \xrightarrow{\cong} (\Lambda^{n-1} N)^* \xrightarrow{(\Lambda^{n-1} f)^*} (\Lambda^{n-1} M)^* \xrightarrow{\cong} M.$$ This terminology is used in arXiv:math/9907114, Section ...


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

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

$$ \begin{split} d(df)&=d(\frac{\partial f}{\partial x^j})\wedge dx^j\\ &=\frac{\partial^2 f}{\partial x^jx^k}dx^k\wedge dx^j\\ &=0 \end{split} $$ since $$ dx^k\wedge dx^j=-dx^j\wedge dx^k $$ and $$ \frac{\partial^2 f}{\partial x^jx^k}=\frac{\partial^2 f}{\partial x^kx^j} $$


1

Here is a slightly more systematic answer than in my comment above. Suppose we have the $(k-1)$-form $\omega=f(\vec{x})\omega_{i_1}$ where $\omega_{i_1}:=dx_{i_2}\wedge dx_{i_3}\wedge\cdots \wedge dx_{x_k}$ has been introduced for convenience. Then the exterior derivative of $\omega$ is $$d\omega=df \omega_{x_i}=\left(\frac{\partial f}{\partial ...



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