New answers tagged

1

How is the interior derivative a derivative? I wouldn't say it is. My background is in Clifford algebra, and that discipline's equivalent of this operation is universally referred to as a product operation, not a derivative operation. What is the geometric content of Hodge duality? Short version: you're finding the orthogonal complement of whatever ...


14

Consider a little block in the fluid with edge lengths $a,b,c$. Velocity at the origin is $\vec{v}=(u,v,w)$. With help of a Taylor series expansion we can approximately write the velocity at $\vec{r} = (a,b,c)$ as: $$ \vec{v}_P = \vec{v}(\vec{r}) = \vec{v}(a,b,c) \approx \vec{v} + \frac{\partial \vec{v}}{\partial x}a + \frac{\partial \vec{v}}{\partial y}b + ...


2

You were on the right track: antisymmetrization over symmetric indices results in zero. To make things extra clear, note that in our equation $\text{(1a)}$, we have $$ \Gamma^\mu_{[\nu\sigma,\rho]} = \Gamma^\mu_{[[\nu\sigma],\rho]} = \Gamma^\mu_{[[(\nu\sigma)],\rho]} = 0, $$ where $\Gamma^\mu_{\nu\sigma,\rho} := \partial_\rho\Gamma^\mu_{\nu\sigma}$. ...


0

You are implicitly using the canonical map $V \to V^{\ast\ast}$ given by $v\mapsto \text{ev}_v$, where $\text{ev}_v$ is the evaluation linear functional on $V^{\ast}$ sending $\phi$ to $\phi(v)$. This is an isomorphism in the finite dimensional case, which you probably also are assuming. This makes $v\in V$ into a map $v:V^\ast\to\Bbb R$ by $v(\psi) = \psi(v)...


3

This seems to be best answered by Lounesto's paper "Marcel Riesz's Work on Clifford Algebras" (see here or here). In what follows: $\bigwedge V=$ the exterior algebra over $V$ $C\ell(Q)=$ the Clifford (geometric) algebra over $V$ w.r.t. the quadratic form $Q$ Note in particular that we always have $C\ell(0)=\bigwedge V$, $0$ being the degenerate ...


2

Your calculation of $e^0$ and $e^1$ is wrong. The easiest way to solve is to find the matrix of the inverse metric tensor: $$(g^{ij}) = \pmatrix{2 & 4 \\ 4 & 10}^{-1} = \frac{1}{\det(g_{ij})}\pmatrix{10 & -4 \\ -4 & 2} = \pmatrix{\frac 52 & -1 \\ -1 & \frac 12}$$ Then $$(e^0)^T = g^{0j}e_j = \frac 52\pmatrix{1 \\ 1} -1\pmatrix{1 \...


3

In euclidean geometry you have a so called affine structure, which essentially states that points are separated by translation vectors. An affine space is a set of points $A$ together with a vector space $V$ and an action of $V$ on $A$ as an abelian group: If $p\in A$ is a point and $v\in V$ is a vector then $p+v\in A$ is another point of $A$, and we can ...


1

I am pretty sure most people as they are learning mathematics think of a vector as an arrow in space. This is useful but it isn't strictly correct. What is a vector then? A vector is an element of a vector space. So the answer to your question is actually fairly simple. An operator can be a vector by being an element of a vector space. I would recommend ...


1

Since your question was about the metric tensor, I'll use that as my example. To answer your question, $g_{ij}$ is a $function$ defined in a neighborhood of your manifold. More specifically, since $g$ is a $(0,2)$-tensor field, we can write $g = \sum_{i,j} g_{ij}\hspace{2pt} dx^i \otimes dx^j$. That is, in a neighborhood, we can expand $g$ in the basis $\{...


2

Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism? Yes and no. In geometric algebra, dual vectors can be computed through Hodge duality. Let $\{u_1, u_2, \ldots, u_n\}$ be an orthogonal basis set for an $n$-dimensional vector space. Let $I$ be their geometric product, which is grade-$n$ due to orthogonality. ...


0

$$(- ik_i \frac{dM_i}{dt} - \frac{1}{2}k_i k_j \frac{dU_{ij}}{dt} - iY_{ij}k_iM_j - Y_{ij}k_iU_{jl}k_l + D_{ij}k_ik_j)P = 0$$ We know $P \ne 0$, real and imagine equal zero separately. $$- \frac{1}{2}k_i k_j \frac{dU_{ij}}{dt} - Y_{ij}k_iU_{jl}k_l + D_{ij}k_ik_j = 0 ~~~~~~~(1)$$ $$- k_i \frac{dM_i}{dt} - Y_{ij}k_iM_j = 0~~~~~~~~(2)$$ for (1), $$- \...


0

This is kind of very late to the party, but if anyone needs this stuff again, in this question Finding basis of isotropic tensors of rank $n$ bases for isotropic tensors up to 8th-order have been computed using Mathematica, such that you can use them for explicit computations.


4

Any finite-dimensional vector space $V$ is isomorphic to a coordinate vector space $\mathbb{R}^n$ which can be thought of as the vector space of functions $\{1,\cdots,n\}\to \mathbb{R}$. Generalizing this, we can use a different index set than $\{1,\cdots,n\}$, for instance we can use a continuous interval as an index set. Then a "coordinate vector" whose ...


0

Here's what I have put together so far, please tell me what you think: For some linear operator L that doesn't change the indices of $h_{\mu\nu}$ $$L(\bar{h}_{\mu\nu})=L(h_{\mu\nu})-L(\frac{1}{2}g_{\mu\nu}(g^{\alpha\beta}h_{\alpha\beta}))$$ $$\bar{\{L(h_{\mu\nu}})\}=L(h_{\mu\nu})-\frac{1}{2}g_{\mu\nu}(Tr[Lh_{\alpha\beta}])$$ Then: $$L(g_{\mu\nu}(g^{\...


2

Fix a basis $\{e_1, \ldots, e_n\}$ of $V$, and consider the dual basis $\{f_1, \ldots, f_n \}$ of $V^\ast$. Then we have a basis $$\{e_1\otimes f_1,\ldots, e_i \otimes f_j, \ldots, e_n \otimes f_n\}$$ for $V \otimes V^\ast$, and the matrix $$A = (a_{ij})$$ is just a way of representing the element $$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \; e_i \otimes f_j \in V \...


1

These resources explaining how to represent multivectors as tensors might be helpful or interesting to you (or those with similar questions): http://www2.ic.uff.br/~laffernandes/teaching/2013.1/topicos_ag/lecture_18%20-%20Tensor%20Representation.pdf https://www.docdroid.net/uwfvUxE/tensor-representation-of-geometric-algebra.pdf.html


1

I'll try to give an answer to this question that doesn't require knowing what a tensor is. Of course it will require some comfort with linear algebra, but I'll even try to keep the multilinear algebra to a minimum. Bye_World's answer in the comments is a good one, although there is slightly more that you can say: a matrix is naturally identified with a rank-...


1

A section of a bundle $E$ (with various properties) is a fancier way of referring to a "smoothly varying" choice of $s_p\in E_p$ (with the same properties) as $p$ varies over $M$. So (1) and (2) are identical. With regard to (2) and (3), we're just using the isomorphism (truly a definition) $\text{Hom}(E,\Bbb R) = E^*$ (where here $E=TM\otimes TM$). Notice ...


1

2. Which types of tensors admit a representation using geometric algebra? On p.4 of this document we can see that multivectors over a given Euclidean space $\mathbb{R}^n$ do not have arbitrarily high grade/order; instead the highest order possible is $n$ (the determinant/volume element). This is because of the two products available in geometric algebra, ...


0

A section of the bundles involved in the second and the third definition can be zero, and not define a metric. After the change in your question, you can analyze the situation in a chart, (since metric are defined on charts then glued with partition of the unity) that is in an open subset $U$ of $R^n$, since the tangent bundle of an open subset of $R^n$ is ...


1

Depends on what context did you study (old school, coordinate-based) differential geometry. Sometimes in physics courses, you have something called tensor calculus, which often just deals with doing tensor calculus in general coordinates in euclidean space. In that case, if you have a coordinate system $(u^1,...,u^n)$, and coordinate basis vectors $\mathbf{...


1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Let $V$ be a real vector space with basis $(\vec{\Basis}_{\mu})_{\mu=1}^{n}$, and assume $V$ is equipped with an inner product $g$ (non-degenerate symmetric bilinear pairing). The author of the notes defines two isomorphisms from $V$ to the dual space $V^{*}$: On page 8,...



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