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對給定黎曼度量的二階張量的跡 (trace) [ edit ]
已知
(
1
,
1
)
{\displaystyle (1,1)}
-張量
F
=
∑
i
k
F
i
k
d
x
i
⊗
∂
∂
x
k
{\displaystyle \textstyle F=\sum _{ik}F_{i}^{k}\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}}}
的跡 (trace)為
tr
(
F
)
=
∑
l
F
l
l
{\displaystyle \textstyle \operatorname {tr} (F)=\sum _{l}F_{l}^{l}}
。
對
(
2
,
0
)
{\displaystyle (2,0)}
-張量
T
{\displaystyle T}
我們定義其對應黎曼度量
g
{\displaystyle g}
的跡 (trace)為:
tr
g
(
T
)
=
def
tr
(
T
♯
)
{\displaystyle \operatorname {tr} _{g}(T){\overset {\underset {\text{def}}{}}{=}}\operatorname {tr} (T^{\sharp })}
,
T
♯
{\displaystyle \textstyle T^{\sharp }}
是
(
1
,
1
)
{\displaystyle (1,1)}
-張量,所以已知其跡 是如何計算。
設
T
=
∑
i
j
T
i
j
d
x
i
⊗
d
x
j
{\displaystyle \textstyle T=\sum _{ij}T_{ij}\,dx^{i}\otimes dx^{j}}
,為
(
2
,
0
)
{\displaystyle (2,0)}
-張量場,將分量
T
i
j
{\displaystyle T_{ij}}
T
i
j
{\displaystyle \textstyle T_{ij}}
第二指數上升讓
(
2
,
0
)
{\displaystyle (2,0)}
-張量變成
(
1
,
1
)
{\displaystyle (1,1)}
-張量,即
T
♯
=
∑
i
k
T
i
k
d
x
i
⊗
∂
∂
x
k
=
∑
i
j
k
g
j
k
T
i
j
d
x
i
⊗
∂
∂
x
k
(
∵
T
i
k
=
∑
j
T
i
j
g
j
k
)
{\displaystyle T^{\sharp }=\sum _{ik}T_{i}^{k}\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}}=\sum _{ijk}g^{jk}T_{ij}\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}}\quad (\because T_{i}^{k}=\sum _{j}T_{ij}g^{jk})}
。
我們有
tr
g
(
T
)
=
def
tr
(
T
♯
)
=
tr
(
∑
i
k
(
∑
j
g
j
k
T
i
j
)
d
x
i
⊗
∂
∂
x
k
)
=
∑
i
(
∑
j
g
j
i
T
i
j
)
=
∑
i
j
g
i
j
T
i
j
{\displaystyle \operatorname {tr} _{g}(T){\overset {\underset {\text{def}}{}}{=}}\operatorname {tr} (T^{\sharp })=\operatorname {tr} (\sum _{ik}(\sum _{j}g^{jk}T_{ij})\,dx^{i}\otimes {\frac {\partial }{\partial x^{k}}})=\sum _{i}(\sum _{j}g^{ji}T_{ij})=\sum _{ij}g^{ij}T_{ij}}
。
tr
g
(
T
)
=
∑
i
j
g
i
j
T
i
j
{\displaystyle \textstyle \operatorname {tr} _{g}(T)=\sum _{ij}g^{ij}T_{ij}}
為對
(
2
,
0
)
{\displaystyle (2,0)}
-張量求跡 (trace)的公式。
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。