# Tag Info

2

[To fix notation: I'll use Greek spacetime indices $\alpha,\beta,\ldots$ running over $0,1,2,3$, and Latin spatial indices $i,j,k$ running over $1,2,3$.] There are four equations in $\partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha } = 0$, depending on which index is missing. A nice way to rewrite it ...

2

This is a variation of physicist Dirac, bra-ket notation. In a more standard bra-ket notation your equation reads $$trace ( |v_4 \rangle \langle v_1 | \otimes |v_3 \rangle \langle v_2 | )$$ Using first $trace ( A \otimes B) = trace (A) trace (B)$, this evaluates nicely to (simply "permute cyclically the kets to the left") $$\langle v_1 | v_4 \rangle ... 3 Since Q is a 3\times3 matrix, Cayley-Hamilton theorem says that Q^3-tQ^2+sQ-rI=0, for some specific scalar (r,s,t). For example, t is the trace of Q, hence t=0 in the present case, and r is the determinant of Q. Hence the identity to prove holds as soon as$$s=-\tfrac12\mathrm{tr}(Q^2). $$The characteristic polynomial x^3-tx^2+sx-r is ... 1 If you have x = x^i\mathbf e_i, and the elements of A are A^i{}_j, then A^Tx must be x^iA^i{}_j\mathbf e_j. The usual rule that each dummy index pair must have one index "up" and one "down" gets reversed because you're specifying the transpose of A. If you want to follow the rule strictly (which you should if your tensors may transform ... 0 Using Einstein summation notation,$$ \begin{align} b^k_\ell = \delta_{i \ell} \delta^{j k} a^i_j. \end{align} $$0 What about B = \delta_{jk} \delta^{il} a^k_l e_i \otimes e^j? 1 A (1,1)-tensor can be thought of as a linear map that sends vectors to vectors; so given a vector X based at p, \nabla\xi(X)=\nabla_X \xi will be another vector based at p, which you should think of as the change in the vector field \xi when you move a small amount in the direction X starting from p. 1 The element xab^{-1} is an element in the localization S^{-1}M where S = R \setminus \{0\}. There is a natural isomorphism M \otimes_R S^{-1}R \simeq S^{-1}M defined by m \otimes \frac{a}{b} \mapsto \frac{ma}{b}. Here S^{-1}R is the ring K. In general writing x_1 \otimes x_2 = x_1x_2 doesn't make sense unless you are invoking some type of ... 1 Here is how I interpret the question: The set of annihilators is Ani(e_1 \wedge e_2 + e_3 \wedge e_4) = \{v \in V\ |\ (e_1 \wedge e_2 + e_3 \wedge e_4) \wedge v = 0\} Note that any element v \in V can be written as: v = \alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \alpha_4 e_4 So v \in Ani(e_1 \wedge e_2 + e_3 \wedge e_4) implies that (e_1 ... 0 Imagine a cube of sponge . Press it with two fingers on two opposite, parallel faces.Meanwhile give a parallel stress on the other sides as well. If you represent that using a tensor, that would be a rank two tensor. Each face has two unit indices. 1 \{e_i\} is a dual basis, i.e.,$$e_i^\ast (e_j)=\delta_{ij}.$$That is any linear map from V to {\bf R} can be written by$$ \sum_{i=1}^4 c_i e_i^\ast$$Note that V\otimes V\ (=M_4({\bf R})) is vector space whose basis is \{ e_i\otimes e_j \} So$$(e=)\ e_1\wedge e_2+e_3\wedge e_4=e_1\otimes e_2 - e_2\otimes e_1+ e_3\otimes e_4 - e_4\otimes e_3 ...

0

CLAIM: $F$ is surjective. For any $n_j \in N_j$, $F(0, \ldots, b_1, \ldots, 0) = n_j$, so by linearity, $F$ is surjective. CLAIM: $F$ is injective. Suppose that $F(n_1, \ldots, n_k) = 0$, so $n_1 + \cdots + n_k = 0$. By (4), $n_1 = \cdots = n_k = 0$, so $(n_1, \ldots, n_k) = (0, \ldots, 0) \in N_1 \times \cdots \times N_k$. Therefore, $F$ is an ...

Top 50 recent answers are included