Hot answers tagged exterior-algebra
8
Short answer:
the exterior derivative acts on differentail forms;
the Lie derivative acts on any tensors and some other geometric objects (they have to be natural, e.g. a connection, see the paper of P. Petersen below);
both the exterior and the Lie derivatives don't require any additional geometric structure: they rely on the differential structure of ...
8
That depends on the convention in your textbook or notes. But I have seen that notation used, so I won't rule it out.
"As $\omega$ is a two-form hence $\omega\wedge\omega\neq 0$" is false. Consider the two form $\mathrm{d}x\wedge \mathrm{d}y$ on $\mathbb{R}^4 = \{(w,x,y,z)| w,x,y,z\in\mathbb{R}\}$. This two form wedged with itself is zero. What you meant ...
6
$$v \otimes w + w \otimes v = (v+w) \otimes (v+w) - v\otimes v - w\otimes w$$
so the latter ideal contains the first one. For field of characteristic not 2, the first one contains the latter one. So in $char \neq 2$ cases they are the same.
I believe that $v \wedge v = 0$ is a condition you always want, which cannot be deduced from your first definition. ...
6
Since I don't have the time to give a super-detailed answer, allow me to just summarize some things that others have said, adding some additional points in the process. Hopefully this will be at least somewhat helpful.
Basic differences:
The exterior derivative and Lie derivative are defined in terms of the structure of a smooth manifold. By contrast, ...
6
The canonical map $\psi\colon V^k \to \bigwedge^k V$ given by $\psi(v_1, \ldots, v_k) = v_1 \wedge \cdots \wedge v_k$ is alternating. If you have an element $g \in (\bigwedge^k V)^*$ then $g \circ \psi\colon V^k \to \mathbf R$ is also alternating, and I believe that this assignment $g \mapsto g \circ \psi$ gives you an inverse to $\Phi$.
This is simpler ...
5
You don't need to choose any particular basis; all you need is the fact that $\wedge V$ is graded and $L_b$ raises the degree by $1$, so if you choose any basis that respects the gradation, then $L_b$ sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish.
5
Let me focus on the difference between Lie derivatives and covariant derivatives. Suppose I have a manifold with a connection $\nabla$ and a point $p$ in the manifold. Let $v$ be a vector field on $M$ and take $\xi \in T_pM$. The point to stress is that $\xi$ is not a vector field (although in practice it is often a vector field evaluated at $p$). We can ...
5
If you want an explicit coboundary, take the boundary of the function $g$ sending $[x^a y^b]$ to $x^{a-2} y^b$ times the image of $a(a-1)/2$ in $\mathbb{F}_2$. It works because the cup square of $\partial / \partial x$ sends $[x^i y^j | x^k y^l]$ to $ikx^{i+k-2}y^{j+l}$, and $dg ( [x^i y^j | x^k y^l]) = x^i y^j g(x^k y^l) - g(x^{i+k}y^{j+l}) + g(x^i y^j) ...
4
We have $y = (y_1, \dots, y_n) = (\phi_1, \dots, \phi_n)$ so
\begin{align*}
dy_1\wedge\dots\wedge dy_n &= d\phi_1\wedge\dots\wedge d\phi_n\\
&= \left(\frac{\partial \phi_1}{\partial x_1}dx_1 + \dots +\frac{\partial \phi_1}{\partial x_n}dx_n\right)\wedge\dots\wedge\left(\frac{\partial \phi_n}{\partial x_1}dx_1 + \dots +\frac{\partial ...
4
This is pointwise. Choose a basis of the 1-forms $\omega_1, \ldots, \omega_n.$ Let $I$ denote any subset of $\{1,2,\ldots,n \}$ containing $k$ elements. then let $$ \omega_I = \omega_{i_1} \wedge \cdots \wedge \omega_{i_k}. $$ Meanwhile, let $I'$ denote the subset consisting of the other $n-k$ indices, that is $$ I \cap I' = \{ \}, \; \; I \cup I' = ...
3
I think that even if it's not written explicitly anywhere, the $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ convention is the most predictable and sensible.
I've never seen the distinction made explicit, since in most circumstances the operation involved is commutative.
I did see somewhere on m.SE someone suggest ...
3
I will keep referring to the Leibniz-rule for exterior derivatives. That is axiom 3 here.
The expression $\rho \wedge \mathrm{d}\rho = 0$ is only true for all one-forms $\rho$ in less than or equal to two dimensions. In dimension $\geq 3$, let $x,y,z$ be the first three of the coordinate functions, we can consider the one form
$$ \rho = x \wedge ...
3
A very short answer:
In finite dimensions and at least in characteristic 0, the equation $$\operatorname{d} \omega(x, y) = \omega(x) - \omega(y) - \omega([x, y])$$ allows you to define $[-,-]: V \wedge V \to V$ if you know $\operatorname{d}: V^* \to V^* \wedge V^*$ and vice versa.
Furthermore, you can prove that conditions $[[x, y], z] + [[y, z], x] + [[z, ...
3
We will use http://en.wikipedia.org/wiki/Exterior_algebra as a reference
Here $$(f \wedge g) (x_1, \dots, x_{k+l}) = \sum_{\sigma \in S_n} sgn(\sigma) f(x_{\sigma (1)}, \dots, x_{\sigma(k)}) g( x_{\sigma(k+1)}, \dots, x_{\sigma(k+l)}) $$
$$(af \wedge bg) (x_1, \dots, x_{k+l}) = \sum_{\sigma \in S_n} sgn(\sigma) a f(x_{\sigma (1)}, \dots, ...
3
Green's Theorem states that, for a simply connected region $D$:
$$\oint_{\partial D} (P dx + Q dy) = \iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right )$$
Choose $P=-y$ and $Q=x$ to get for the area $A(D)$
$$A(D) = \frac{1}{2} \oint_{\partial D} (-y \,dx + x \,dy)$$
For an ellipse, $x=a \cos{t}$, $y=b \sin{t}$, and we ...
3
This is a variant of the polarization trick:
$$
v \otimes w + w \otimes v = (v + w) \otimes (v + w) - v \otimes v - w \otimes w
$$
Let $I$ be your ideal generated by all $v \otimes w + w \otimes v$ and let $J$ be the ideal generated by all $w \otimes w$.
The polarization identity shows that $I$ is contained in $J$. Of course, you can go the other way ...
3
Not sure if this is a rigorous explanation but an intuitive way to think of this: The idea of quotiening out by the ideal $\{ v \otimes w+ w \otimes v | v,w \in W\}$ is that we impose the relation
$$v \wedge w = - w\wedge v$$
in the quotient where $v \wedge w$ is the image of $v \otimes w$ under the canonical map. Now what happens when we quotient out by ...
3
To add to Qiaochu's answer, often one can "explain" the existence of a grading on a mathematical object by giving a representation of the "circle group" (analytically, $S^1$, algebraically, the multiplicative group). The idea is that a representation $V$ of the circle group (here I'll use the analytic language) canonically decomposes into Fourier modes: the ...
3
These definitions are the same. In fact, there is a third, more common definition that they are equivalent to. Namely, if we take a local coordinate system $(x^{1}, \cdots, x^{n})$, we get $dx^1 , \cdots , dx^n$ as a basis for the cotangent space of a point in the chart. If we take a multi-index $I = (i_{1}, \cdots, i_{k})$, then we can consider a $k$-form ...
2
This is essentially by the definition you wrote. It is assumed that $x_1, \ldots, x_n$ is a basis, and by definition $dx_1, \ldots, dx_n$ are the dual basis. Thus if you expand out $v_j=a_{j1}v_1+\cdots + a_{jn}v_n$ in this basis, then $dx_i(v_j)=a_{ji}$.
Also in this basis $(v_1, \ldots, v_n)=(a_{ij})$. So since ...
2
Your $\wedge^k A^m$ is just the polarization:
$\wedge^k A^m = {N \choose m} \wedge^k(\underbrace{A,\dots,A}_m,1,\dots,1)$,
the right-hand side refers to $\wedge^k$ as a $k$-linear form. As we know, $\wedge^N$ is the determinant, so $\wedge^N A^m$ are just the coefficients of the characteristic polynomial. Those are elementary symmetric functions of the ...
2
One problem with your approach is that your basis is too generic, which is inconvenient for computations. Try some special basis: e.g. Combine $a_1e_1 \wedge e_2 + a_2 e_1 \wedge e_3 + a_3 e_1\wedge e_4$ as $e_1 \wedge (a_1e_2 + a_2e_3 + a_3e_4)$. Then let $e_2' = a_1e_2 + a_2e_3 + a_3e_4$ we can replace the sum of first three terms as $e_1 \wedge e_2'$. ...
2
You can prove this by proving the contrapositive. Suppose $\omega$ is not decomposable.
That means that $\omega$ necessarily must be the sum of two (and only two) decomposable terms, like $\omega = x \wedge y + z \wedge w$. Consider what happens if you add a term like $y \wedge z$. You should realize that you can lump that into one of the two terms and ...
2
I think it might be better to leave the wedges in: $dx\wedge dt = - dt\wedge dx$. Otherwise this quickly leads to confusion. Now to you question.
For one thing: If you do it this way, the determinant pops out naturally under change of bases. And this is essential for integration, which is what these things were designed to do (remember the change of ...
2
As a general remark, you are asking how to recognize perfect wedges in $\bigwedge^2 V$ via equations in higher exerior powers. This is related to the Plucker relations for Grassmanians.
E.g. a non-zero perfect wedge $v_1\wedge v_2$ in $\bigwedge^2 V$ corresponds to a $2$-plane in $V$ (the span of $v_1$ and $v_2$), are you are asking whether or not the ...
2
1) The vector derivative, $\partial$
In geometric calculus, one deals in not just vector fields but multivector fields--fields that associated oriented planes, volumes, or other types of primitives to each point. These multivector fields are differentiated by an operator denoted $\partial$. It can act on multivector fields in either of two ways. On a ...
2
You might consider forming the 2-vector $\sum_{i,j} (x_i e_i) \wedge e_j$. You can then consider the action of the 2-form on the 2-vector just by matching up similar components: the component of $e_1^* \wedge e_2^*$ multiplies the component of $e_1 \wedge e_2$, and so on. Only these corresponding components multiply and contribute to the scalar result.
...
2
Let $$\alpha = \sum_{i = 1}^{n} f_{i}\ \textrm{d}x_{i}$$
be a closed one-form with functions $f_{i}$ smooth and homogeneous of degree $p$.
Since $\alpha$ is closed, $\textrm{d}\alpha = 0$ and so
$$\sum_{j = 1}^{n}\sum_{i = 1}^{n}\frac{\partial f_{i}}{\partial x_{j}} \textrm{d}x_{j}\wedge \textrm{d}x_{i} = 0, $$
$$\Rightarrow \frac{\partial f_{i}}{\partial ...
2
You can define $\omega_f$ by
$$\omega_f(u, v) = f(u)(v),$$
i.e. you take the action of $f(u) \in V^\ast$ on the vector $v \in V$. Since $f$ is a skew map, we have that
$$\omega_f(v, u) = f(v)(u) = f^\ast(u)(v) = -f(u)(v) = -\omega_f(u,v),$$
so $\omega_f$ is skew-symmetric. Presumably $f$ is also linear, from which the bilinearity of $\omega_f$ follows. Hence ...
2
If your elements commute with eachother, then there is no need for an ordering in the case of finite sums/products. In the non-commutative case things are more complicated.
Anyhow, IMO there is no need for an ordering if the sum/product doesn't depend on the order. And this covers many non-commutative cases too. Otherwise, it is clear that one should ...
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