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4

For a given basis $\{e_a\}$, $g_{ab}=g(e_a,e_b)$ denotes the components of the metric $g$ with respect to the given basis. So (at each point of the manifold) $g$ is a bilinear form, and in a given basis it is represented by a matrix. While matrices generally do not commute, their components are just numbers and do commute. In your example $R_{bacd}$ is just ...


4

$v\otimes w$ is either of the following: The rank 1 linear mapping $V^*\to V$ given by $\alpha\mapsto v.\langle \alpha,w\rangle$ where $\langle\;,\;\rangle:V^*\times V\to \mathbb F$ is the duality. This is the easiest to visualize. The decomposable bilinear form $V^*\times V^*\to \mathbb F$ given by $(\alpha,\beta)\mapsto \langle\alpha,v\rangle ...


3

In chapter 2, when talking about Riemann Normal Coordinates, Carroll shows that you can cunningly pick coordinates such that at a given point the metric has its standard form $$\begin{pmatrix} -1 & & &\\ & 1& & \\ & &1 & \\ & & & 1\\ \end{pmatrix}$$ and so that the first derivatives all vanish $$\partial_\rho ...


2

Okay, I think I understand enough of your question to give an answer. No And that is because two flat metrics on the same 2D manifold do not have to be conformal. To wit, let $\mathbb{T}^2$ denote the torus. The metrics with line elements $$ \mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 $$ and $$ \mathrm{d}s^2 = \mathrm{d}x^2 + 2 \mathrm{d} y^2 $$ ...


2

It is useful to look at this situation from a more general setting. Suppose we have a hypersurface $S$ in a $(n+1)$-dimensional Riemannian manifold $M$. (In your case $M = \mathbb{R}^3$ with the standard (Euclidean) metric). By a hypersurface we mean a sumbanifold of codimension $1$, so $\dim S = n$. The unit normal vector $N$ at each point $p \in S$ is ...


2

Let $V$ be a real, finite-dimensional vector space $V$ and $$T: V^* \otimes V \to \mathbb{R}$$ a $(1, 1)$-tensor. Now, given any $\lambda \in V^*$, we denote by $$T(\lambda, \,\cdot\,)$$ the map $V \to \mathbb{R}$ defined by $$X \mapsto T(\lambda, X).$$ Since $T$ is multilinear, the map $T(\lambda, \,\cdot\,)$ is linear, but linear maps $V \to \mathbb{R}$ ...


2

You can't multiply by $\epsilon^{ijk}$ because there's already a $k$ in $B_k$ and indices should not appear more than twice.


2

Yes. If you want to do it explicitly, just multiply both sides with the inverse metric to raise both indices.


1

It's very common to write symmetric products using juxtaposition with no product symbol. But you should never omit the wedge symbol in a wedge product. Maybe you're getting confused with the notation for integrals, where, for example $\int f\,dx^1\wedge dx^2$ is defined to mean $\int f(x^1,x^2)\, dx^1dx^2$. The $dx^1dx^2$ in the latter expression is not a ...


1

That equation is incorrect; I'm not used to this notation but I think it's actually $$e_{ijk} e_{lmk} = \delta_{il} \delta_{jm} + \delta_{im} \delta_{jl}.$$ In fact a tensor with this property does not exist in any dimension at least $2$. The reason is that it would imply the existence of a pair of linear maps $f : V \otimes V \to V$ and $g : V \to V ...


1

No. For a two dimensional example, take $$ T = \begin{pmatrix} x & 0 \\ -y & 0 \end{pmatrix} $$ We have $$ \partial_x T_{1j} + \partial_y T_{2j} = (0,0) $$ but $$ \partial_x T_{i1} + \partial_y T_{i2} = (1,0) $$ Observe that for symmetric and antisymmetric tensors, $\partial^i T_{ij} = 0 \iff \partial^j T_{ij} = 0$. So the same is true of ...


1

The inclusion $T^k(M) \to T(M)$ induces a map $$ T^k(M)/I \cap T^k(M) \to T(M)/I = S(M) $$ which is obviously injective. Your definition of $S^k(M)$ is exactly the submodule of $S(M)$ which is the image of this map (since obviously $T^k(M) \to S(M)$ is trivial on $I \cap T^k(M)$). Therefore, the two definitions give canonically isomorphic modules, and it is ...


1

For $v_i\in V,w_i^*\in W^*$, we can see $\sum_i v_i\otimes w_i^*$ as an element of $Hom(W,V)$, by setting : $$ \forall w\in W, \qquad (\sum_i v_i\otimes w_i^*)(w)= \sum_i w_i^*(w)v_i. \quad (A) $$ Let $(j_1^*,\cdots,j_n^*)$ be the dual base of $B_W=(j_1,\cdots,j_n)$. (a) The matrix of $\:e_i\otimes j_l^*$ in the bases you gave $(B_W \: and \: (e_1, ...


1

Note that $x^T A^T A x = (Ax)^T (Ax) = \lVert Ax \rVert^2$, so this is expressible as a norm. We expect the derivative to be something like $2A^T Ax$, which is what you have in the line $$ \frac{\partial L}{\partial x_l} = \delta_{\alpha l} + \lambda a_{jl}(a_{ji} x_i + a_{jk}x_k): $$ changing the indices, this is $$ \frac{\partial L}{\partial x_l} = ...


1

An alternative approach, if $A$ is invertible: First consider the case when $A=I$. Then the problem is to maximize $u^Tx$ subject to $x^T x\le r^2$, where $u$ is a fixed vector ($e_\alpha$ in your case). By Cauchy-Schwarz, the solution is $x=ru/|u|$. For general invertible $A$, the problem is to maximize $u^Tx$ subject to $x^TA^TAx\le r^2$. Let $y=Ax$. ...


1

Since no one has answered yet, I'm assuming your still having trouble with it. The trick for your question was to realise $$\begin{align} \implies [(\vec A \times \vec B) \times (\vec C \times \vec D)]_{i} &= [-(\vec B \times \vec A) \times (\vec C \times \vec D)]_{i} \\ &= [(\vec C \times \vec D) \times (\vec B \times \vec A)]_{i} \end{align}$$ ...


1

After some thinking I decided to add another answer to this question. The reason for doing this is that I want to keep my previous answer for the reference, while adding a comment or expanding the previous answer would make the texts unreadable. Let us begin with refining the statement of the original problem. Suppose we have an open set $U \subseteq ...


1

This is defined on page $13$ of the same book. A tensor field $Y \in T^k_l(M)$ is a $C^{\infty}(M)$-linear map $Y : (\Omega^1(M))^l\times (\mathfrak{X}(M))^k \to C^{\infty}(M)$. The trace of $Y$ is the tensor field $\operatorname{tr}Y \in T^{k-1}_{l-1}(M)$ satisfying $$\operatorname{tr}Y(\omega_1, \dots, \omega_{l-1}, V_1, \dots, V_{k-1}) = ...


1

The first thing to notice is in your last equation, the symmetry of $g^{\mu\alpha}$ means that $$ g^{\mu\alpha} \partial_{\mu} g_{\nu \alpha} = g^{\mu\alpha} \partial_{\alpha} g_{\mu\nu}, $$ so those two terms cancel and we are left with proving that $$ (\partial_{\nu} \log{g} =) \, \frac{\partial_{\nu} g}{g} = g^{\mu\alpha}\partial_{\nu} g_{\mu\alpha} ...


1

When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Yes. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is ...



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