Recent content by yifli

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    Why is the differential being onto equivalent to it not being zero?

    the differential of f(x)=x^2 is 2x, so f'(x)=0 means x=0. Now what?
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    Why is the differential being onto equivalent to it not being zero?

    I have difficulty understanding the following Theorem If U is open in ℝ^2, F: U \rightarrow ℝ is a differentiable function with Lipschitz derivative, and X_c=\{x\in U|F(x)=c\}, then X_c is a smooth curve if [\operatorname{D}F(\textbf{a})] is onto for \textbf{a}\in X_c; i.e., if \big[...
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    What is the definition of Lie derivatives?

    I read the book again and found out it's just the notation they use: for any differentiable function f defined about x and any tangent vector \xi they set \xi(f)=D_\varphi(f) where \varphi \in \xi (they define a tangent vector as an equivalence class), so D_{\varphi_x}f=X(x)f @quasar987: The...
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    What is the definition of Lie derivatives?

    Let \varphi be a one-parameter group on a manifold M, and let f be a differentiable function on M, the derivative of f with respect to \varphi is the defined as the limit: \lim_{t\to 0} \frac{\varphi^*_t[f]-f}{t}(x)=\lim_{t\to 0}\frac{f\circ \varphi_x(t)-f\circ...
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    Convergence of a sequence of points on a manifold

    If I choose the chart (U_1,\phi_1), where U_1 is the lower semi-sphere, then the first condition is not met because the north pole is in the upper semi-sphere; also, since the points converge to the north pole, how do you find an integer N such that x_k \in U_1 for k>N?
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    Convergence of a sequence of points on a manifold

    I have a question regarding the following definition of convergence on manifold: Let M be a manifold with atlas A. A sequence of points \{x_i \in M\} converges to x\in M if there exists a chart (U_i,\phi_i) with an integer N such that x\in U_i and for all k>N,x_i\in U_i \phi_i(x_k)_{k>N}...
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    Diagonalization of symmetric bilinear function

    According to duality principle, a bilinear function \theta:V\times V \rightarrow R is equivalent to a linear mapping from V to its dual space V*, which can in turn be represented as a matrix T such that T(i,j)=\theta(\alpha_i,\alpha_j). And this matrix T is diagonalizable, i.e...
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    What is the Name for the Linear Mapping f* and How is it Proven?

    Thanks. Still, can anyone show me a proof of the f*(dw)=d(f*w)
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    Line Integral w/ Respect to Arc Length & x/y

    There are line integral with respect to arc length and line integral with respect x/y. I know \int_C Pdx+Qdy is useful to calculate the work. When do we need the line integral with respect to arc length?
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    What is the Name for the Linear Mapping f* and How is it Proven?

    Is there a name for the linear mapping f^*:\wedge^k(R^{m}_{f(p)}) \rightarrow \wedge^k(R^{n}_{p}) where f is a differentiable mapping from R^n \rightarrow R^m. When k is 1, f* is called the adjoint of f. But what about k > 1? Also can someone show me a proof of f^*(d\omega)=d(f^*\omega)...
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    Are Order 1 Alternating Tensors Different from Normal Tensors?

    I don't understand how order 1 alternating tensors fit the definition of alternating tensors. I think the order has to be at least 2 for the definition to make sense because only then can we talk about permutation. So I think there's no difference between order 1 alternating tensors and...
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    What determines the orientation of a vector space?

    A non-zero alternating tensor w splits the bases of V into two disjoint groups, those with \omega(v_1,\cdots,v_n)>0 and those for which \omega(v_1,\cdots,v_n)<0. So when we speak of the orientation of a vector space, we need to say the orientation with respect to a certain tensor, correct?
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    Proof that the rationals are dense

    To prove the rationals are dense in the reals, you need to prove for any real number, there exists a rational number which is arbitrarily close to the real number. I don't see this in your proof.
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    The mapping to alternating tensors

    I'm wondering why 1/k! is needed in Alt(T), which is defined as: \frac{1}{k!}\sum_{\sigma \in S_k} \mbox{sgn}\sigma T(v_{\sigma(1)},\cdots,v_{\sigma(k)}) After removing 1/k!, the new \mbox{Alt}, \overline{\mbox{Alt}}, still satisfies...
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    Unique volume element in a vector space

    So given a fixed orientation, it's impossible for \det(A) to be -1? What does it mean for \det(A) to be -1?
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