Найти значение $F(v,f)$ тензора

$F=e^{1}\otimes e_{2}+e^{2}\otimes (e_{1}+3e_3)\in\mathbb{T}_{1}^{1}\left(V\right),$ 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 $F$:

$$F(v,f)=e^{1}\left(e_{1}+5e_{2}+4e_{3}\right)\otimes e_{2}\left(e^{1}+e^{2}+e^{3}\right)+e^{2}\left(e_{1}+5e_{2}+4e_{3}\right)\otimes (e_{1}\left(e^{1}+e^{2}+e^{3}\right)+3e_{3}\left(e^{1}+e^{2}+e^{3}\right))=$$

$$e^{1}\left(e_{1}\right)\otimes e_{2}\left(e^{2}\right)+e^{2}\left(5e_{2}\right)\otimes (e_{1}\left(e^{1}\right)+3e_{3}\left(e^{3}\right))=1\otimes 1+5\otimes 4=21.$$

Найти значение тензора $A\otimes B-B\otimes A\in\mathbb{T}_{5}^{0}\left(V\right)$ от набора

$A=e_{1}\otimes e_2+e_2\otimes e_3+e_2\otimes e_2$

$B=e_1\otimes e_2\otimes \left(e_1-e_3\right)$

$v_{1}=e_1,v_2=e_1+e_2,v_3=e_2+e_3,v_4=v_5=e_2$

$B=e_1\otimes e_2\otimes \left(e_1-e_3\right)$

$A=e_1\left(e_1\right)\otimes e_2\left(e_1+e_2\right)+e_2\left(e_1\right)\otimes e_3\left(e_1+e_2\right)+e_2\left(e_1\right)\otimes e_2\left(e_1+e_2\right)=$

$1\otimes 1+0\otimes 0+0\otimes 1=1$

$B=e_1\left(e_2+e_3\right)\otimes e_2\left(e_2\right)\otimes \left(e_1\left(e_2\right)-e_3\left(e_2\right)\right)=0\otimes 1\otimes 0=0$

тогда $A\otimes B(v)=0$

$B=e_1\left(e_1\right)\otimes e_2\left(e_1+e_2\right)\otimes \left(e_1\left(e_2+e_3\right)-e_3\left(e_2+e_3\right)\right)=1\otimes 1\otimes \left(-1\right)=-1$

$A=e_1\left(e_2\right)\otimes e_2\left(e_2\right)+e_2\left(e_2\right)\otimes e_3\left(e_2\right)+e_2\left(e_2\right)\otimes e_2\left(e_2\right)=0\otimes 1+1\otimes 0+1\otimes 1=1$

тогда $B\otimes A(v)=-1$

следовательно $\left(A\otimes B-B\otimes A\right)\left(v\right)=0-(-1)=1$

$A=e_1\otimes e_2+e_2\otimes e_3+e_3\otimes e_1$

$B=\sum e_i\otimes e_j\otimes e_k$

$v_1=e_1+e_2,v_2=e_3+e_2,v_3=e_1+e_3,v_4=v_5=e_2$

$A=e_1\left(e_1+e_2\right)\otimes e_2\left(e_3+e_2\right)+e_2\left(e_1+e_2\right)\otimes e_3\left(e_3+e_2\right)+e_3\left(e_1+e_2\right)\otimes e1\left(e_3+e_2\right)=$

$1\otimes 1+1\otimes 1+0\otimes 0=2$

$B=\sum e_i\otimes e_j\otimes e_k=e_i\left(e_1+e_3\right)\otimes e_j\left(e_2\right)\otimes e_k\left(e_2\right)=2$

$A=e_1\left(e_2\right)\otimes e_2\left(e_2\right)+e_2\left(e_2\right)\otimes e_3\left(e_2\right)+e_3\left(e_2\right)\otimes e_1\left(e_2\right)=0\otimes 1+1\otimes 0+0\otimes 0=0$

$B=\sum e_i\otimes e_j\otimes e_k=e_i\left(e_1+e_2\right)\otimes e_j\left(e_3+e_2\right)\otimes e_k\left(e_1+e_3\right)=8$

Тогда

$A\otimes B-B\otimes A=2\otimes 2-8\otimes 0=4$

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

$v=e_{1}+2e_{2}+3e_{3}+4e_{4},f=e^{1}-e^{4}$

$F=3\sum e^{i}\otimes e^{j}\otimes e^{k}\otimes e_{l}\otimes e_{m}$

$F=3\sum e^{i}\left(e_{1}+2e_{2}+3e_{3}+4e_{4}\right)\otimes e^{j}\left(e_{1}+2e_{2}+3e_{3}+4e_{4}\right)\otimes e^{k}\left(e_{1}+2e_{2}+3e_{3}+4e_{4}\right)\otimes e_{l}\left(e^{1}-e^{4}\right)\otimes e_{m}\left(e^{1}-e^{4}\right)=0$

Найти координаты с индексами 1, 2, 3, 3, 3 произведений $A\otimes B$ и $B\otimes A$ тензоров

$A=e1\otimes e2+e3\otimes e3$

$B=\det\left(v1,v2,v3\right)$

$A\otimes B\left(e1,e2,e3,e3,e3\right)=A\left(e1,e2\right)\otimes B\left(e3,e3,e3\right)=A\left(e1,e2\right)\otimes 0=0$

$B\otimes A\left(e1,e2,e3,e3,e3\right)=B\left(e1,e2,e3\right)\otimes A\left(e3,e3\right)=B\left(e1,e2,e3\right)\otimes \left(e1\left(e3\right)\otimes e2\left(e3\right)+e3\left(e3\right)\otimes e3\left(e3\right)\right)=1\otimes (0+1)=1$

Найти координаты:

$\widetilde{t_{21}^{1}}$ 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:

$t_{i}^{j}=(e_{1}+3e_{2}-e_{3})\otimes (e^{1}-2e^{3}+3e^{4})-(e_{1}+e_{3})\otimes (e^{1}-3e^{3}+e^{4})$;

$t_{i}^{i}=(e_{1})\otimes (e^{1})+(-e_{3})\otimes (-2e^{3})-(e_{1})\otimes (e^{1})-(e_{3})\otimes (-3e^{3})=1+2-1+3=5$