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0

There is some value of $\sigma_{22}$ for which t=(0,0,0) on some (unknown) plane. Since $[t]=[\sigma][n]$, this is equivalent to the condition that the homogeneous linear equation $[\sigma][n]=[0]$ has a non-trivial solution. This occurs when $\det[\sigma]=0$. From the component form, evaluating the determinant gives you $0 + 4-\sigma_{22}+4=0$, so the ...


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Hint: Take a linear combination $L=k_1a+k_2b+k_3c+k_4d$ and evaluate at a vector $X$ for which $a(X)=1$ , $b(X)=0$ , $c(X)=0$ and $d(X)=0$.


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In 2) you have how much is the value of $T$ at the arguments $v_{\mu_1},...,V_n$.


1

Let's be very general. Assume we are taking the metric tensor of the given point $g = g_{ij}(x)$. Assume everything in arbitrary coordinate system. Assume $M$ to be a arbitrary differentiable manifold, for an arbitrarily given metric tensor $g$. For a given vector x, y: $$ x\cdot y = g_{ij}(x)x^i y^j $$ For a commutative inner product of vectors implies a ...


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$\DeclareMathOperator\Sq{Sq}$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

$$U_{ij} = g_{ij} + \epsilon_{ijk}u^k\tag{1}$$ $$ (U^{-1})^{jl} = Ag^{jl} + Bu^ju^l + C\epsilon^{jlm}u_m.\tag{2}$$ $$\delta_i^l=U_{ij} (U^{-1})^{jl} = g_{ij}Ag^{jl} + g_{ij}Bu^ju^l+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{3}$$ $$\delta_i^l= g_{ij}Ag^{jl} + B(g_{ik}u^k)(g^{lm}u_m)+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{4}$$ $$\delta_i^l= A g_{i}^l + ...


2

Yes. Because $$I=\epsilon_{ijk}A^{ijk}=-\epsilon_{jik}A^{ijk}=-\epsilon_{jik}A^{jik}=-I$$. So $I = 0$


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

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 $\epsilon_{ijk4}A^{jk}=-\epsilon_{ij4k}A^{jk}$.


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 ...


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.


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Hint: Take, for example the mixed $(2,1)$-tensors $$T={T_{i_1i_2}}^{j_1}dx^{i_1}\otimes dx^{i_2}\otimes\frac{\partial}{\partial x^{j_1}},$$ as tri-indexed linear combination on the basis $dx^{i_1}\otimes dx^{i_2}\otimes\frac{\partial}{\partial x^{j_1}}$. These basis, as tri-linear maps $T_pM\times T_pM\times T_p^*M\to{\mathbb{R}}$, work via ...


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

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

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 ...


0

Let $\epsilon \in \bigwedge^n E$. Then for any $A: E \mapsto E$, the determinant $\det A$ obeys $$A^n(\epsilon) = (\det A) \epsilon$$ Now do this for $A = I + f$, and let $X = x_1 \wedge x_2 \wedge \ldots \wedge x_n$: $$\det(I + f) \epsilon(X) =\epsilon [ (I+f)(x_1) \wedge (I+f)(x_2) \wedge \ldots \wedge (I+f)(x_n)]$$ Now, suppose the vectors $x_i$ form ...


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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 ...


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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 ...


0

$\blacktriangleright$ For k=1 , it follows by definition. $\blacktriangleright$ For $2\leq k\leq \lfloor \frac{n}{2}\rfloor$, we will show that $\omega:=e_{1}\wedge...\wedge e_{ \lfloor \frac{n}{2}\rfloor}+e_{\lfloor \frac{n}{2}\rfloor+1}\wedge...\wedge e_{ 2\lfloor \frac{n}{2}\rfloor}$ is non-decomposable (the floor notation was added to work for odd and ...


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 $\frac{\partial F}{\partial x^k}=\frac{1}{2F}\frac{\partial g_{ab}}{\partial x^k}\dot{x}^a\dot{x}^b$


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 ...


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 ....


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Have you thought, once the inner product is defined, to rewrite it in more universal terms? Work first with $V\otimes V$, and see if you can do it.


0

I can't really make sense of your idea for line integrals. Using parallel transport you can more or less reduce this to the case of integrating a tensor on an interval-but how do you do that? Normally integration should output a number, and for this it really seems you need a 1-form. Given a metric, it's true that you can do better: you can integrate ...


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

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

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 ...


0

If we have some vector $u$, then its components $(u^j)^n_{j=1}$ in the basis $\alpha$ can be found by applying the dual basis $\alpha^*$ to $u$. $$u^j=v^j(u)$$ So when we change from $\alpha$ to $\beta$ the components change by $$u^j=v^j(u)\longmapsto w^j(u)=(S\mathbf{v})^j(u)=\sum_i{\mu^j}_iu^i$$ So the components of vectors change by left multiplication ...


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 ...


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 ...


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 ...


0

(cross-posting a variation of my answer from this) Both co-variant and contra-variant vectors are just vectors (or more generaly 1-order tensors). Furthermore they are vectors which relate to the same underlying space (for example euclidean space or generally, a manifold) Furthermore they relate to the same space but in different but dual ways (as such ...



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