Kronecker delta: in $3-$ dimension $delta_ii=3$

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My teacher said that in $3-$ dimension $delta_ii=3$, but why?



Kronecker delta's definition:



$$delta_ij=begincases0& textif; ineq j \ 1 & textif; i=j endcases$$



According to the definition $delta_ij(jto i)=delta_ii=^?1$



My teacher said since we do not know the $i$ so there is a secret sum in $delta$ so we get $delta_11+delta_22+delta_33=3$. Why? In the definition I cannot see this implication.




notation index-notation




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    Looks like you are expected to apply Einstein's summation convention. That is standard in some contexts.
    – Jyrki Lahtonen
    Jul 21 at 20:57














up vote
1
down vote

favorite












My teacher said that in $3-$ dimension $delta_ii=3$, but why?



Kronecker delta's definition:



$$delta_ij=begincases0& textif; ineq j \ 1 & textif; i=j endcases$$



According to the definition $delta_ij(jto i)=delta_ii=^?1$



My teacher said since we do not know the $i$ so there is a secret sum in $delta$ so we get $delta_11+delta_22+delta_33=3$. Why? In the definition I cannot see this implication.




notation index-notation




share|cite|improve this question















  • 2




    Looks like you are expected to apply Einstein's summation convention. That is standard in some contexts.
    – Jyrki Lahtonen
    Jul 21 at 20:57












up vote
1
down vote

favorite









up vote
1
down vote

favorite











My teacher said that in $3-$ dimension $delta_ii=3$, but why?



Kronecker delta's definition:



$$delta_ij=begincases0& textif; ineq j \ 1 & textif; i=j endcases$$



According to the definition $delta_ij(jto i)=delta_ii=^?1$



My teacher said since we do not know the $i$ so there is a secret sum in $delta$ so we get $delta_11+delta_22+delta_33=3$. Why? In the definition I cannot see this implication.




notation index-notation




share|cite|improve this question











My teacher said that in $3-$ dimension $delta_ii=3$, but why?



Kronecker delta's definition:



$$delta_ij=begincases0& textif; ineq j \ 1 & textif; i=j endcases$$



According to the definition $delta_ij(jto i)=delta_ii=^?1$



My teacher said since we do not know the $i$ so there is a secret sum in $delta$ so we get $delta_11+delta_22+delta_33=3$. Why? In the definition I cannot see this implication.





notation index-notation





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 21 at 20:54









user2312512851

1,139521




1,139521







  • 2




    Looks like you are expected to apply Einstein's summation convention. That is standard in some contexts.
    – Jyrki Lahtonen
    Jul 21 at 20:57












  • 2




    Looks like you are expected to apply Einstein's summation convention. That is standard in some contexts.
    – Jyrki Lahtonen
    Jul 21 at 20:57







2




2




Looks like you are expected to apply Einstein's summation convention. That is standard in some contexts.
– Jyrki Lahtonen
Jul 21 at 20:57




Looks like you are expected to apply Einstein's summation convention. That is standard in some contexts.
– Jyrki Lahtonen
Jul 21 at 20:57










1 Answer
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2
down vote



accepted










It's just the trace, written with the Einstein summation convention:



$$delta_ii = delta_11 + delta_22 + delta_33 = 3$$



$$delta_ij = beginpmatrix
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
endpmatrix
$$



Einstein summation convention



When the same index is repeated inside an expression, it means summation over the repeated index:



$$A_ii = sum_i = 1^n A_ii = A_11 + A_22 + ldots + A_nn$$






share|cite|improve this answer





















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    1 Answer
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    active

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    1 Answer
    1






    active

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    oldest

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    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    It's just the trace, written with the Einstein summation convention:



    $$delta_ii = delta_11 + delta_22 + delta_33 = 3$$



    $$delta_ij = beginpmatrix
    1 & 0 & 0 \
    0 & 1 & 0 \
    0 & 0 & 1
    endpmatrix
    $$



    Einstein summation convention



    When the same index is repeated inside an expression, it means summation over the repeated index:



    $$A_ii = sum_i = 1^n A_ii = A_11 + A_22 + ldots + A_nn$$






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      It's just the trace, written with the Einstein summation convention:



      $$delta_ii = delta_11 + delta_22 + delta_33 = 3$$



      $$delta_ij = beginpmatrix
      1 & 0 & 0 \
      0 & 1 & 0 \
      0 & 0 & 1
      endpmatrix
      $$



      Einstein summation convention



      When the same index is repeated inside an expression, it means summation over the repeated index:



      $$A_ii = sum_i = 1^n A_ii = A_11 + A_22 + ldots + A_nn$$






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        It's just the trace, written with the Einstein summation convention:



        $$delta_ii = delta_11 + delta_22 + delta_33 = 3$$



        $$delta_ij = beginpmatrix
        1 & 0 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1
        endpmatrix
        $$



        Einstein summation convention



        When the same index is repeated inside an expression, it means summation over the repeated index:



        $$A_ii = sum_i = 1^n A_ii = A_11 + A_22 + ldots + A_nn$$






        share|cite|improve this answer













        It's just the trace, written with the Einstein summation convention:



        $$delta_ii = delta_11 + delta_22 + delta_33 = 3$$



        $$delta_ij = beginpmatrix
        1 & 0 & 0 \
        0 & 1 & 0 \
        0 & 0 & 1
        endpmatrix
        $$



        Einstein summation convention



        When the same index is repeated inside an expression, it means summation over the repeated index:



        $$A_ii = sum_i = 1^n A_ii = A_11 + A_22 + ldots + A_nn$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 21 at 20:57









        Von Neumann

        16k72443




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