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3

Hint: Use Lemma $4.6$ (ii) (i.e. the formula displayed here) to deduce that $$(\nabla_X g)(Y, Z) = \nabla_Xg(Y, Z) - g(\nabla_XY, Z) - g(Y, \nabla_XZ).$$


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This is a matter of convention that differs between literatures and depending on what you're doing. If you're going to use raised indices in a meaningful way, then I'd expect to see the basis written as $$ B = \{ u^i \}_{i=1}^{\infty} $$ with the index up on vectors. Then your second option for $K$ would be correct. Since you wrote the vectors in the ...


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There is nice characterisation of tensors which $Alt(T)=0.$ Thm. Let $V$ be a vector space over $\mathbb{R}$ (without any additional assumptions). Consider subspace $N^n(V)$ of $\otimes^nV$ generated by elements $v_1\otimes\dots\otimes v_n$ such that $v_i=v_j$ for at least one pair $i\neq j.$ We have that $$\ker(Alt)=N^n(V).$$ Equivalently. For every ...


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It seems that it is Einstein's sum notation. So you have actually $A = (a_{ij})_{1\le i, j \le 3}$ with $$ a_{ij} = \sigma_{ij} + \sum_{k=1}^3 \left( \sigma_{ik} w_{kj} - w_{ik} \sigma_{kj} \right), $$ or $$ A = \Sigma + \Sigma W - W \Sigma, $$ with $\Sigma = (\sigma_{ij})_{1\le i, j \le 3}$ and $W = (w_{ij})_{1\le i, j \le 3}$.


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I don't know about the specific equation in hand but a common way to represent matrix elements is to using $\sigma_{ij}$ to mean the element of a matrix, say $\Sigma$ at the $i$th row and the $j$th column. Then $\sigma_{ik}\omega_{kj}$ is the product of the element of $\Sigma$ matrix at the $i$th row and the $k$th column and that of $\Omega$ matrix at the ...


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Using $x$ and $y$ as parameters makes perfect sense. This means you are using the diffeomorphism $\Phi(x,y) = (x,y,\sin x+\sin y)$ to cover the whole surface with a coordinate patch. Pushing forward the standard vector fields $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ from $\mathbb{R}^2$ to the surface gives a basis of the tangent space, ...


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Actually, matrix multiplication can be seen as a bilinear map $$K^{n\times k}\times K^{k\times m}\rightarrow K^{n\times m},$$ where $K$ is the ground field. Such a bilinear map can be expressed as a tensor of order 3 of format $(nk)\times (km)\times (nm)$ (any bilinear map $K^{\ell_1}\times K^{\ell_2}\rightarrow K^{\ell_3}$ can be represented by a tensor of ...


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The induced metric is just the Euclidean metric on $\mathbb R^{n+1}$, i.e. $g_{ij} = \delta_{ij}$, but restricted to act on vectors tangent to $S^n$. You need to choose a system of $n$ coordinates parametrizing $S^n$ alone (or an $n$-frame tangent to $S^n$) if you want to get an $n \times n$ matrix of components for $g$.


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This is a terrific question. I will try to answer all of your concerns. In general, you should be aware of a few things. First, when we say things like $g_{ij}$, $i$ and $j$ are just placeholders for integers. So it doesn't matter what letters are in the subscript, they mean the same thing. Second, not all tensors can be represented as matrices. Only tensors ...


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We could write your tensor as $$ (a_1,a_2) \otimes (b_1,b_2) \otimes (c_1, c_2) $$ Or, depending on your notation, perhaps $$ (b_1,b_2) \otimes (c_1, c_2) \otimes (a_1,a_2) $$ One way to check that this tensor is rank one is to note that one matrix is a multiple of the other, and that each matrix is rank one.


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Yes, it makes sense to talk about Christoffel symbols in flat spacetime. Every coordinate system has associated Christoffel symbols. On Minkowski spacetime in the standard coordinates, the Christoffel symbols are all zero. But in different coordinates (e.g., spherical coordinates), they will not be zero. The Christoffel symbols contain information about the ...


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Killing fields generate isometries and the scalar curvature is isometry-invariant, so moving along a Killing field does not change the curvature. More precisely: since the scalar curvature is just a function, the covariant derivative $V^m \nabla_m R$ is equal to the Lie derivative $$\mathcal L_V R = \frac{d}{dt}\Big|_{t=0}(R\circ \Phi^V_t)$$ where $\Phi^V$ ...


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I think I see another argument, but I'm not sure it is valid, because everyone is using different terminology and I'm a bit confused by the profusion of terms. I know "geodesic coordinates" as coordinates in which the Christoffel symbols of the second kind vanish at a point. Their derivatives do not vanish here. Various people talk about "normal ...


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I think it's true. If $G$ is your metric tensor, then, using the formula for the derivative of the inverse, your expression is zero if $tr(G^{-1}G_{x_j}G^{-1}G_{x_k})=tr(G^{-1}G_{x_k}G^{-1}G_{x_j})$, where $G_{x_i}$ is the derivative of $G$ wrt $x_i$. Let $A=G^{-1}G_{x_k}$ and $B=G^{-1}G_{x_j}$. Since $G$ is symmetric, so are $A$ and $B$, and your expression ...


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Note that the indices of the metric tensor are raised in the left-most expression, but lowered in the middle expression. Recall that $$ g^{ij}\,g_{js} = \delta^i_s $$ Take the partial derivative with respect to $q^k$ of both sides and you'll find where the negative sign went. As for the right-most expression, it's simply the middle one where the time term ...


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There are often different ways of viewing the same sort of tensor, and people will often not bother to make the distinction. Natural isomorphisms between the operations that proudce the various tensors spaces justify this identification. In fact some notations for working with tensors will not even make a choice as to which viewpoint is being taken. Here is ...



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