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3

Notice that $\dfrac{\partial x_p}{\partial x_q}=\delta_{pq}$. Then $$\dfrac{\partial H}{\partial x_m} = \sum_{i,j,k=1}^n\ J_{i,j,k}(\delta_{im} x_j x_k+x_i \delta_{jm} x_k+x_i x_j \delta_{km})=\\\sum_{j,k=1}^n\ J_{m,j,k} x_j x_k+\sum_{i,k=1}^n\ J_{i,m,k} x_i x_k+\sum_{i,j=1}^n\ J_{i,j,m} x_i x_j.$$

2

Let us compute $dF(X)$ using directional derivatives: for $M\in\mathbb R^{2n\times 2n}$ we have $$dF(X)M=\lim_{t\to0}\frac{F(X+tM)-F(X)}{t}=\lim_{t\to0}\frac{(X+tM)A(X+tM)^{-1}-XAX^{-1}}{t}.$$ Now for $|t|$ small we can use the expression $$(X+tM)^{-1}=X^{-1}-tX^{-1}MX^{-1}+t^2P,$$ so that $$... 2 You really need to take a CW structure on \bar X which is inherited by a CW structure on X. This means a priori your CW structure is fixed during the whole construction, and in the end you show the independence of the resulting homology of the chosen structure (as you do in cellular homology as well). Fix a CW structure on X. The covering space ... 2 In \Bbb{R}^4, certainly, the volume form is dV = dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4. However, confined to the sphere S^3 in \Bbb{R}^4, you need a 3-form. (It is perhaps easier to see this for the sphere S^2 in \Bbb{R}^3, where what is wanted is for areas, not volumes.) 2 The fact that it is a sum of vectors is not relevant. Let v = \sum_{i=1}^ne_i, then the left hand side becomes \Delta(v, v - e_2, v - e_3, \dots, v - e_n). By linearity in the second argument, we have$$\Delta(v, v - e_2, v - e_3, \dots, v - e_n) = \Delta(v, v, v - e_3, \dots, v - e_n) + \Delta(v, - e_2, v - e_3, \dots, v - e_n)$$As \Delta is ... 2 The metric specifies a canonical isomorphism between V and V^*--and therefore an invertible map g' : V \to V^*. Consider some T: V \times V^* \to \mathbb R. Now consider T': V \times V \to \mathbb R such that T'(A, B) = T(A, g'(B)). That is what we're doing when we raise or lower indices. We might have some multilinear function that takes ... 2 How can you choose x_i^*,x_j^* among the elements of the dual basis? Well, they can either be different, which gives you \binom{n}{2} elements, or you can have x_i^*=x_j^*, giving you n more possibilities, so$$\dim\{x^*_i \otimes x^*_j + x^*_j \otimes x^*_i \mid x^*_i,x^*_j \in V^*\} = \binom{n}{2} + n.$$Now,$$\binom{n}{2} + \left(\binom{n}{2} + ...

2

There are two relevant operations you want to be familiar with on tensors. First, given a $(m,n)$ tensor $T$ and an $(k,l)$ tensor $S$, you can construct their tensor product $T \otimes S$ which will be a $(m + k, n + l)$ tensor. Invariantly, $$(T \otimes S)(\varphi^1, \ldots, \varphi^m, \varphi^{m+1}, \ldots, \varphi^{m + k}, v_1, \ldots, v_n, v_{n+1}, ... 2 Symmetry: If (A_1,B_1)\sim (A_2,B_2) this means A_2=S^{-1}A_1S and B_2 = S^{-1}B_1S. Now you can choose S'=S^{-1} and you get (A_2,B_2)\sim (A_1,B_1) Reflexivity Choose S=I the identity than you get that A_1=S^{-1}A_1 S and B_1=S^{-1}B_1 S. So you obtain (A_1,B_1)\sim (A_1,B_1) Transitive This means: Assume (A_1,B_1)\sim (A_2,B_2) ... 1 If M is finitely generated, then clearly M/A_+M is finitely generated, since it is a quotient of M. Conversely suppose that M/A_+M is finitely generated, let B=\{b_1,\dots,b_n\} be a finite set of homogeneous generators of cardinal n, and let \phi:A^n\to M be the unique A-linear map such that \phi(e_i)=b_i for each i\in\{1,\dots,n\}. The ... 1 Hint: the form dV is defined on the Lie algebra su(2) of SU(2) you can represent the elements of su(2) by complex 2\times 2-matrices and use the trace and Lie bracket. su(2)=\pmatrix{ia & -c+id\cr c+id & -ia}a,b,c\in R. thus su(2) SU(2)=\pmatrix{a & -\bar b\cr b & \bar a}, a,b\in C \mid a\mid^2+\mid b\mid^2=1 thus it is a ... 1 Let me explain supposing that m=2 first. If you consider the set of all bilinear maps V\times V\to\Bbb R, with V={\rm span}\{b_i\}, one should also consider that the effect of taking a pair of basic duals \beta^j such that \beta^j({b_i})=\delta^j{}_i. With pair of \beta^i,\beta^j one gets$$\beta^i\otimes\beta^j:V\times V\to\Bbb R given by ...

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