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==對給定黎曼度量的二階張量的跡(trace)==
==對給定黎曼度量的二階張量的[[ 跡]] (trace)==
已知若<math>F=\sum_{ik}F_i^k\,dx^i\otimes \frac{\partial}{\partial x^k}</math>是<math>(1,1)</math>-張量，則其跡(trace)為<math>\operatorname{tr}(F)=\sum_l F_l^l</math>。
已知若<math>F=\sum_{ik}F_i^k\,dx^i\otimes \frac{\partial}{\partial x^k}</math>是<math>(1,1)</math>-張量，則其[[ 跡]] (trace)為<math>\operatorname{tr}(F)=\sum_l F_l^l</math>。
對二階張量<math>X</math>我們定義其對應黎曼度量<math>g</math>的跡(trace)為：
對二階張量<math>X</math>我們定義其對應黎曼度量<math>g</math>的[[ 跡]] (trace)為：
:<math>\operatorname{tr}_g(X):=\operatorname{tr}(X^\sharp)</math>，
:<math>\operatorname{tr}_g(X):=\operatorname{tr}(X^\sharp)</math>，
<math>X^\sharp</math>是<math>(1,1)</math>-張量，所以其跡是有定義的。
<math>X^\sharp</math>是<math>(1,1)</math>-張量，所以其[[ 跡]] 是有定義的。
設<math>X=\sum_{ij}X_{ij}\,dx^i\otimes dx^j</math>，為 <math>(2,0)</math>-張量場，將第二指數上升讓<math>(2,0)</math>-張量變成<math>(1,1)</math>-張量即，
設<math>X=\sum_{ij}X_{ij}\,dx^i\otimes dx^j</math>，為 <math>(2,0)</math>-張量場，將第二指數上升讓<math>(2,0)</math>-張量變成<math>(1,1)</math>-張量即，
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對給定黎曼度量的二階張量的跡 (trace) 已知若
F
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{\displaystyle F=\sum _{ik}F_{i}^{k}\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}}}
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1
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{\displaystyle (1,1)}
跡 (trace)為
tr
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{\displaystyle \operatorname {tr} (F)=\sum _{l}F_{l}^{l}}
對二階張量
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{\displaystyle X}
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{\displaystyle g}
跡 (trace)為：
tr
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{\displaystyle \operatorname {tr} _{g}(X):=\operatorname {tr} (X^{\sharp })}
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{\displaystyle X^{\sharp }}
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{\displaystyle (1,1)}
跡 是有定義的。
設
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{\displaystyle X=\sum _{ij}X_{ij}\,dx^{i}\otimes dx^{j}}
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{\displaystyle (2,0)}
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{\displaystyle (2,0)}
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{\displaystyle (1,1)}
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{\displaystyle X^{\sharp }=\sum _{ijk}g^{jk}X_{ij}\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}}}
我們有
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{\displaystyle \operatorname {tr} _{g}(X):=\operatorname {tr} (X^{\sharp })=\operatorname {tr} (\sum _{ik}(\sum _{j}g^{jk}X_{ij})\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}})=\sum _{i}(\sum _{j}g^{ji}X_{ij})=\sum _{ij}g^{ij}X_{ij}}
注意：雖然這裡是上升第二指數(raising the second index)，不過對第一指數上升也會得到相同結果。
