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47.2. Найти значение $F(v,f)$ тензора
$F=e^{1}\otimes e_{2}+e^{2}\otimes (e_{1}+\lyxmathsym{З}\lyxmathsym{е}_{3})\in\mathbb{T}_{1}^{1}\left(V\right),$ где $v=e_{1}+5e_{2}+4e_{3}$, $f=e^{1}+e^{2}+e^{3}$.
Для того чтобы вычислить $F(v,f)$, нужно подставить значения $v$ и $f$ в $F$:
$F(v,f)=e^{1}\left(e_{1}+5e_{2}+4e_{3}\right)\otimes e_{2}\left(e^{1}+e^{2}+e^{3}\right)+e^{2}\left(e_{1}+5e_{2}+4e_{3}\right)\otimes (e_{1}\left(e^{1}+e^{2}+e^{3}\right)+3e_{3}\left(e^{1}+e^{2}+e^{3}\right))=$
$e^{1}\left(e_{1}\right)\otimes e_{2}\left(e^{2}\right)+e^{2}\left(5e_{2}\right)\otimes (e_{1}\left(e^{1}\right)+3e_{3}\left(e^{3}\right))=1\otimes 1+5\otimes 4=21$
47.3. Найти значение тензора $A\otimes B-B\otimes A\in\mathbb{T}_{5}^{0}\left(V\right)$ от набора
a) $A=e_{1}\otimes e2+e2\otimes e3+e2\otimes e2$
$B=e1\otimes e2\otimes \left(e1-e3\right)$
$v_{1}=e1,v2=e1+e2,v3=e2+e3,v4=v5=e2$
$B=e1\otimes e2\otimes \left(e1-e3\right)$
$A=e1\left(e1\right)\otimes e2\left(e1+e2\right)+e2\left(e1\right)\otimes e3\left(e1+e2\right)+e2\left(e1\right)\otimes e2\left(e1+e2\right)=$
$1\otimes 1+0\otimes 0+0\otimes 1=1$
$B=e1\left(e2+e3\right)\otimes e2\left(e2\right)\otimes \left(e1\left(e2\right)-e3\left(e2\right)\right)=0\otimes 1\otimes 0=0$
тогда $A\otimes B(v)=0$
$B=e1\left(e1\right)\otimes e2\left(e1+e2\right)\otimes \left(e1\left(e2+e3\right)-e3\left(e2+e3\right)\right)=1\otimes 1\otimes \left(-1\right)=-1$
$A=e1\left(e2\right)\otimes e2\left(e2\right)+e2\left(e2\right)\otimes e3\left(e2\right)+e2\left(e2\right)\otimes e2\left(e2\right)=0\otimes 1+1\otimes 0+1\otimes 1=1$
тогда $B\otimes A(v)=-1$
следовательно $\left(A\otimes B-B\otimes A\right)\left(v\right)=0-(-1)=1$
б) $A=e1\otimes e2+e2\otimes e3+e3\otimes e1$
$B=\sum ei\otimes ej\otimes ek$
$v1=e1+e2,v2=e3+e2,v3=e1+e3,v4=v5=e2$
$A=e1\left(e1+e2\right)\otimes e2\left(e3+e2\right)+e2\left(e1+e2\right)\otimes e3\left(e3+e2\right)+e3\left(e1+e2\right)\otimes e1\left(e3+e2\right)=$
$1\otimes 1+1\otimes 1+0\otimes 0=2$
$B=\sum ei\otimes ej\otimes ek=ei\left(e1+e3\right)\otimes ej\left(e2\right)\otimes ek\left(e2\right)=2$
$A=e1\left(e2\right)\otimes e2\left(e2\right)+e2\left(e2\right)\otimes e3\left(e2\right)+e3\left(e2\right)\otimes e1\left(e2\right)=0\otimes 1+1\otimes 0+0\otimes 0=0$
$B=\sum ei\otimes ej\otimes ek=ei\left(e1+e2\right)\otimes ej\left(e3+e2\right)\otimes ek\left(e1+e3\right)=8$
Тогда
$A\otimes B-B\otimes A=2\otimes 2-8\otimes 0=4$
47.4. Найти значение $F\left(v,v,v,f,f\right)$ тензора $F\in\mathbb{T}_{3}^{2}\left(V\right)$, если все координаты тензора $F$ равны 3, и
$v=e_{1}+2e_{2}+3e_{3}+4e_{4},f=e^{1}-e^{4}$
$F=3\sum e^{i}\otimes e^{j}\otimes e^{k}\otimes e_{l}\otimes e_{m}$
$F=3\sum e^{i}\left(e_{1}+2e_{2}+3e_{3}+4e_{4}\right)\otimes e^{j}\left(e_{1}+2e_{2}+3e_{3}+4e_{4}\right)\otimes e^{k}\left(e_{1}+2e_{2}+3e_{3}+4e_{4}\right)\otimes e_{l}\left(e^{1}-e^{4}\right)\otimes e_{m}\left(e^{1}-e^{4}\right)=0$
47.6. Найти координаты с индексами 1, 2, 3, 3, 3 произведений $A\otimes B$ и $B\otimes A$ тензоров
$A=e1\otimes e2+e3\otimes e3$
$B=\det\left(v1,v2,v3\right)$
$A\otimes B\left(e1,e2,e3,e3,e3\right)=A\left(e1,e2\right)\otimes B\left(e3,e3,e3\right)=A\left(e1,e2\right)\otimes 0=0$
$B\otimes A\left(e1,e2,e3,e3,e3\right)=B\left(e1,e2,e3\right)\otimes A\left(e3,e3\right)=B\left(e1,e2,e3\right)\otimes \left(e1\left(e3\right)\otimes e2\left(e3\right)+e3\left(e3\right)\otimes e3\left(e3\right)\right)=1\otimes (0+1)=1$
47.7. Найти координаты:
a) $\widetilde{t_{21}^{1}}$ тензора $e^{1}\otimes e^{2}\otimes (e_{1}+e_{2})\in T_{2}^{1}(V)$ в базисе
$(\widetilde{e_{1}},\widetilde{e_{2}})=(e_{1},e_{2})\left(\begin{array}{cc}
1 & 1
2 & 3
\end{array}\right)$
$(e_{1},e_{2})=(\widetilde{e_{1}},\widetilde{e_{2}})\left(\begin{array}{cc}
1 & 1
2 & 3
\end{array}\right)^{-1}=(\widetilde{e_{1}},\widetilde{e_{2}})\left(\begin{array}{cc}
3 & -1
-2 & 1
\end{array}\right)$
$(e^{1},e^{2})=(\widetilde{e^{1}},\widetilde{e^{2}})\left(\begin{array}{cc}
1 & 1
2 & 3
\end{array}\right)$
$\widetilde{t_{jk}^{i}}=\left(\widetilde{e^{1}}+2\widetilde{e^{2}}\right)\otimes \left(\widetilde{e^{1}}+3\widetilde{e^{2}}\right)\otimes (3\widetilde{e_{1}}-2\widetilde{e_{2}}-\widetilde{e_{1}}+\widetilde{e_{2}})=\left(\widetilde{e^{1}}+2\widetilde{e^{2}}\right)\otimes \left(\widetilde{e^{1}}+3\widetilde{e^{2}}\right)\otimes (2\widetilde{e_{1}}-\widetilde{e_{2}})$
$\widetilde{t_{21}^{1}}=\left(\widetilde{e^{1}}+2\widetilde{e^{2}}\right)\otimes \left(\widetilde{e^{1}}+3\widetilde{e^{2}}\right)\otimes (2\widetilde{e_{1}}-\widetilde{e_{2}})\left(\widetilde{e_{2}},\widetilde{e_{1}},\widetilde{e^{1}}\right)=$
$\left(\widetilde{e^{1}}\left(\widetilde{e_{2}}\right)+2\widetilde{e^{2}}\left(\widetilde{e_{2}}\right)\right)\otimes \left(\widetilde{e^{1}}\left(\widetilde{e_{1}}\right)+3\widetilde{e^{2}}\left(\widetilde{e_{1}}\right)\right)\otimes (2\widetilde{e_{1}}\left(\widetilde{e^{1}}\right)-\widetilde{e_{2}}\left(\widetilde{e^{1}}\right))=$
$\left(0+2\right)\otimes \left(1+3\cdot0\right)\otimes (2-0)=2\otimes 1\otimes 2=4$
47.13. Найти полную свертку тензоров:
a) $t_{i}^{j}=(e_{1}+3e_{2}-e_{3})\otimes (e^{1}-2e^{3}+3e^{4})-(e_{1}+e_{3})\otimes (e^{1}-3e^{3}+e^{4})$;
$t_{i}^{i}=(e_{1})\otimes (e^{1})+(-e_{3})\otimes (-2e^{3})-(e_{1})\otimes (e^{1})-(e_{3})\otimes (-3e^{3})=1+2-1+3=5$