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<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,
:<math>X^\sharp=\sum_{ijk}g^{jk}X_{ij}\,dx^i\otimes \frac{\partial}{\partial x^k}</math>。
:<math>X^\sharp=\sum_{ijk}g^{jk}X_{ij}\,dx^i\otimes \frac{\partial}{\partial x^k}</math>。
我們有
我們有
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對給定黎曼度量的二階張量的跡(trace)
已知若
a
=
∑
i
j
a
i
j
d
x
i
⊗
∂
∂
x
j
{\displaystyle a=\sum _{ij}a_{i}^{j}\,dx^{i}\otimes {\frac {\partial }{\partial x^{j}}}}
是
(
1
,
1
)
{\displaystyle (1,1)}
-張量,則其跡(trace)為
tr
(
a
)
=
∑
i
a
i
i
{\displaystyle \operatorname {tr} (a)=\sum _{i}a_{i}^{i}}
。
對二階張量
X
{\displaystyle X}
我們定義其跡(trace)為:
tr
g
(
X
)
:=
tr
(
X
♯
)
,
{\displaystyle \operatorname {tr} _{g}(X):=\operatorname {tr} (X^{\sharp }),}
X
♯
{\displaystyle X^{\sharp }}
是
(
1
,
1
)
{\displaystyle (1,1)}
-張量,所以其跡是有定義的。
設
X
=
∑
i
j
X
i
j
d
x
i
⊗
d
x
j
{\displaystyle X=\sum _{ij}X_{ij}\,dx^{i}\otimes dx^{j}}
,為
(
2
,
0
)
{\displaystyle (2,0)}
-張量場,將第二指數上升讓
(
2
,
0
)
{\displaystyle (2,0)}
-張量變成
(
1
,
1
)
{\displaystyle (1,1)}
-張量即,
X
♯
=
∑
i
j
k
g
j
k
X
i
j
d
x
i
⊗
∂
∂
x
k
{\displaystyle X^{\sharp }=\sum _{ijk}g^{jk}X_{ij}\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}}}
。
我們有
tr
g
(
X
)
:=
tr
(
X
♯
)
=
tr
(
∑
i
k
(
∑
j
g
j
k
X
i
j
)
d
x
i
⊗
∂
∂
x
k
)
=
∑
i
(
∑
j
g
j
i
X
i
j
)
=
∑
i
j
g
i
j
X
i
j
{\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。