# Tag Info

4

You need to be careful about what space you're taking the adjoint in. In the space of all square-integrable $(0,2)$-tensors, your calculation correctly shows that the adjoint of $-\text{div}$ is indeed $\nabla$. I think you want to be working with the space of symmetric trace-free $(0,2)$-tensors - $K$ maps into this space (possibly after tweaking the ...

3

Per my comment (which had a typo and should have said $b_jb_m$) \begin{align} \varepsilon_{ijk}a_ib_j\hat{e}_k\cdot\varepsilon_{lmn}a_lb_m\hat{e}_n &= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m(\hat{e}_k\cdot\hat{e}_n)\\ &= \varepsilon_{ijk}\varepsilon_{lmn}a_ib_ja_lb_m\delta_{kn}\\ &= \varepsilon_{ijk}\varepsilon_{lmk}a_ib_ja_lb_m\\ &= ...

3

Suppose $V$ is a vector space of dimension $n$ over some field, and consider $\alpha\in\Lambda^{n-1}V$ nonzero. If $v_1,\dots,v_n$ is a basis of $V$, then there are scalars $\lambda_1,\dots,\lambda_n$ such that $$\alpha=\sum_{i=1}^n\lambda_i\cdot v_1\wedge\dots\wedge\widehat{v_i}\wedge\dots\wedge v_n$$ Where ...

2

Evaluating $\nabla_\mu(X^\lambda g_{\lambda \nu})$ by the Leibniz rule for derivatives gives $$(\nabla_\mu X^\lambda) g_{\lambda \nu}+X^\lambda (\nabla_\mu g_{\lambda \nu}).$$ But since the covariant derivative of the metric vanishes the second term is zero, so we do indeed have $$(\nabla_\mu X^\lambda)g_{\lambda \nu} = \nabla_\mu(X^\lambda g_{\lambda ... 2 Yes. Because$$I=\epsilon_{ijk}A^{ijk}=-\epsilon_{jik}A^{ijk}=-\epsilon_{jik}A^{jik}=-I$$. So I = 0 2 You mention "the tensor notation of a determinant" in the comments. I don't know quite how your definitions have been given to you, but I imagine that you already know something like this:$$\epsilon_{pmn}a_{pk}a_{mi}a_{nj}=\epsilon_{kij}\operatorname{det}(a).$$Then in the case a is a rotation matrix, we know \operatorname{det}(a)=1, so we have ... 2 Your first equation is wrong: You left out the Levi-Civitta symbol coming from the middle cross product (and thus end up with an expression with no free indices, which you know can't be right). The correct starting point is$$ \epsilon_{krp} (\epsilon_{ijk}a_ib_j)\epsilon_{mnr}a_mc_n) $$which simplifies as follows:$$ \begin{array}{c} \epsilon_{mnr} ...

2

If you lower the indices and consider $$M_{\mu \nu} = g_{\gamma \mu}g_{\delta \nu}M^{\gamma \delta},$$ then $M_{\mu \nu}$ is a two form (by equation two), is co-closed (first equation) and closed (third equation). Thus you are looking for a harmonic two form. (I am treating these as Riemannian manifolds, so I hope I did not misunderstand the concepts).

1

$\DeclareMathOperator\Sq{Sq}$Yes, that's what it means here. The algebra $\mathcal{A}_2$ is the quotient of the free tensor algebra generated by $\Sq^1, \Sq^2\dots$ quotiented by the ideal generated by the Adem relations. Similarly for $\mathcal{A}_p$, $p>2$. However be careful with the admissible thing. What Hatcher shows is that every element of the ...

1

@Marra: Avirus gave a nice explanation of the Schouten-Nijenhuis bracket (useful in Poisson geometry). If you want just to know the Lie derivative of the exterior product $\mathscr{L}_X(Y\wedge Z)$, then you can start from the definition, for any tensor field $T$: $$\mathscr{L}_X T=\frac{d}{dt}\Big|_{t=0}(\exp tX)^*T$$ where $\exp tX$ the local flow of $X$. ...

1

I realize this is an old question, but I was searching for a related problem and I found a simple solution. First, some theory: Let $E_1 = (y,0)$ and $E_2 = (0,y)$ be an orthonormal basis for $T_{(x,y)} \mathbb{H}^2$ the connection form $\omega_{12}$ can be calculated like any connection form in a conformal manifold with ruler $g$, by: $$\omega_{12} = g_y ... 1 Take F^2=g_{ab}\dot{x}^a\dot{x}^b. So 2F\frac{\partial F}{\partial x^k}=\frac{\partial g_{ab}}{\partial x^k}\dot{x}^a\dot{x}^b. Hence \frac{\partial F}{\partial x^k}=\frac{1}{2F}\frac{\partial g_{ab}}{\partial x^k}\dot{x}^a\dot{x}^b 1 Please verify the following calculation.$$ A^{ab}B_{ab} = A^{ab}(-B_{ba}) = -A^{ab}B_{ba} = -A^{ba}B_{ba} = -A^{ab}B_{ab} $$Each step is either a simple algebraic manipulation or uses the assumed properties of A and B. This calculation implies what you want. 1 To clarify my remark in comments: Using any summation index more than twice renders the summation convention is insensible. This becomes obvious if you write in terms of \Sigma's:$$(\vec{a}\times \vec{b})^2=\left(\sum_{j,k=1}^3\epsilon_{ijk}a_j b_k\right)^2\neq \sum_{j,k=1}^3(\epsilon_{ijk}a_j b_k)^2.$$What is correct is$$(\vec{a}\times ...

1

I am using prior knowledge that this is an approach to eigenvalue decomposition to formulate my answer. Let $$R = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix},$$ where $q_1$ is an eigenvector of the matrix $A$. Then $$RAR^T = \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix}^T A \begin{bmatrix} q_1 \ \ \tilde{R} \end{bmatrix} = \begin{bmatrix} q_1 \ \ ... 1 Given your background, it is useful to remember that a manifold is a space that is almost Euclidean in the neigbourhood of each point. This means that locally tensor calculus on manifolds is not that different to working with curvilinear coordinates on euclidean spaces and most of your intuitions from working with curvilinear coordinates should carry over. ... 1 Hint: If each of the e^{*}_{i_{1}}\otimes...\otimes e^{*}_{i_{k}} is orthogonal to each other and normalized, we should have \langle e^{*}_{i_{1}}\otimes...\otimes e^{*}_{i_{k}}, e^{*}_{j_{1}}\otimes...\otimes e^{*}_{j_{k}} \rangle = \delta_{i_1j_1} \cdots \delta_{i_kj_k} Now, by ...., you can extend this definition to .... 1 This is totally correct. In your final line, the term "\boldsymbol{B}^{T}(\boldsymbol{x}):(\nabla \boldsymbol{B}^{T}(\boldsymbol{x}))" is a bit ambiguous since (\nabla \boldsymbol{B}^{T}(\boldsymbol{x})) has three indices and it's not clear which of them the indices of \boldsymbol{B}^{T}(\boldsymbol{x}) are being contracted with. 1$$U_{ij} = g_{ij} + \epsilon_{ijk}u^k\tag{1} (U^{-1})^{jl} = Ag^{jl} + Bu^ju^l + C\epsilon^{jlm}u_m.\tag{2}\delta_i^l=U_{ij} (U^{-1})^{jl} = g_{ij}Ag^{jl} + g_{ij}Bu^ju^l+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{3}\delta_i^l= g_{ij}Ag^{jl} + B(g_{ik}u^k)(g^{lm}u_m)+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{4}\delta_i^l= A g_{i}^l + ...

1

You can use the following definition of the cross product $$a \times b = \epsilon_{ijk} a_j b_k \hat{e}_i$$ So your second cross product $(a \times b) \times (a \times c) =$ is $$\epsilon_{ijk} a_j b_k \hat{e}_i \times \epsilon_{lmn} a_m c_n \hat{e}_l \\ = \epsilon_{rst} (\epsilon_{ijk} a_j b_k \hat{e}_i \cdot \hat{e}_s)(\epsilon_{lmn} a_m c_n ... 1 Well, you can take the equation$$ g_{ij}g^{jk}=\delta_i^k $$as a definition of g^{jk} and establish its transformation properties. This (tensor) equation should be valid in any basis, so$$ g'_{ij}{g'}^{jk}={\delta'}_i^k = {\delta}_i^k $$The change of basis formula for the covariant components is$$ g'_{ij} = \frac{\partial x_p}{\partial ...

1

Note that the permutations corresponding to $(i,j,k,4)$ and $(i,j,4,k)$ differ only by a transposition of the last two indices. Consequently, they have different parity and so their Levi-Civita symbols have opposite signs (assuming they do not vanish, of course). Hence the correct statement is $\epsilon_{ijk4}A^{jk}=-\epsilon_{ij4k}A^{jk}$.

1

If you fix one of the indices of $\varepsilon_{ijkl}$ to be $4$ you get $\pm\varepsilon_{ijk}$ depending on wheather you fix an odd or an even positioned index. So $\varepsilon_{ijk4}=\varepsilon_{ijk}$ but $\varepsilon_{ij4k}=-\varepsilon_{ijk}$. To see why the signs come out this way, notice that when substitute $1,2,3$ for $i,j,k$ you get: ...

1

Let me make my life easier by just proving this for $E = \mathbb R^n$; I'm going to use the standard basis $e_1, \ldots, e_n$ for that, and the dual basis $\phi_1, \ldots, \phi_n$, where $\phi_i(v) = e_i \cdot v$, that's defined by the usual inner product. I'm also going to show that there are functionals $\alpha_i$ such that $$\alpha = r \alpha_1 \wedge ... 1 I will start again, the trouble I spotted right away with your initial calculation is that you have used \mu as a dummy index of summation whereas it is apparently free from the initial expression:$$ \eta^{\mu \nu} F_{\alpha \beta, \nu} F^{\alpha \beta} $$Ok, so, to raise the derivative index, you can just use the existing metric in the expression ... 1 This doesn't use the symmetry of S, but you can use the following identity:$$\epsilon_{ijk}\epsilon_{mnp}=\delta_{im}\delta_{jn}\delta_{kp}+\delta_{in}\delta_{jp}\delta_{km}+\delta_{ip}\delta_{jm}\delta_{kn}-\delta_{im}\delta_{jp}\delta_{kn}-\delta_{ip}\delta_{jn}\delta_{km}-\delta_{in}\delta_{jm}\delta_{kp}$$to get$$(A \times B)\cdot S\cdot(C \times ...

1

Consider a small volume element $$\{(s_1,s_2,s_3)\in \mathbb{R}^3: x_1 \leq s_1 \leq x_1 + \Delta x_1,x_2 \leq s_2 \leq x_2 + \Delta x_2,x_3 \leq s_3 \leq x_3 + \Delta x_3\}.$$ The surface (stress) force acting on that element in the $i$ direction consists of contributions from each of three pairs of faces where coordinates $s_1$, $s_2$ and $s_3$ are ...

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