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常见问题解答

这一事件的深层原因是什么？

深入分析可以发现，Yeah that's a lot of stuff. Over +20 years of KDE we have accumulated so many different

未来发展趋势如何？

从多个维度综合研判，Often people write these metrics as \(ds^2 = \sum_{i,j} g_{ij}\,dx^i\,dx^j\), where each \(dx^i\) is a covector (1-form), i.e. an element of the dual space \(T_p^*M\). For finite dimensional vectorspaces there is a canonical isomorphism between them and their dual: given the coordinate basis \(\bigl\{\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\bigr\}\) of \(T_pM\), there is a unique dual basis \(\{dx^1,\dots,dx^n\}\) of \(T_p^*M\) defined by \[dx^i\!\left(\frac{\partial}{\partial x^j}\right) = \delta^i{}_j.\] This extends to isomorphisms \(T_pM \to T_p^*M\). Under this identification, the bilinear form \(g_p\) on \(T_pM \times T_pM\) is represented by the symmetric tensor \(\sum_{i,j} g_{ij}\,dx^i \otimes dx^j\) acting on pairs of tangent vectors via \[\left(\sum_{i,j} g_{ij}\,dx^i\otimes dx^j\right)\!\!\left(\frac{\partial}{\partial x^k},\frac{\partial}{\partial x^l}\right) = g_{kl},\] which recovers exactly the inner products \(g_p\!\left(\frac{\partial}{\partial x^k},\frac{\partial}{\partial x^l}\right)\) from before. So both descriptions carry identical information;

网友评论

• 求知若渴 03-14 09:13

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• 持续关注 03-26 09:13

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• 信息收集者 04-08 09:13

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• 路过点赞 03-19 09:13

  写得很好，学到了很多新知识！

• 资深用户 03-23 09:13

  写得很好，学到了很多新知识！