Kronecker Delta Expressions











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I am trying to understand the Kronecker Delta and want to clarify. Considering the definition of the Kronecker Delta and assuming $i=j=k$ for the following situations:



I know that $delta _j^i delta _i^j $ is equal to $N$ where $N$ is the dimension of space. Would this mean that $delta _j^i delta _k^j delta _i^k $ would also be equal to $N$?



Similarly, since $delta _i^i delta _j^j $ is equal to $N^2$, would $delta _i^i delta _j^j delta _k^k $ be equal to $N^3$ ?



Last, is $delta _i^j delta _j^k $ just equal to $delta _i^k = 1 $ ?






vectors tensors kronecker-symbol kronecker-delta







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  • @AlvinJin This should be an answer not a comment
    – Eddy
    Sep 25 at 20:51










  • If $i=k$, wouldn't the $delta_i^k$ be equal to 1, by definition though?
    – Cave Johnson
    Sep 25 at 20:53















up vote
2
down vote

favorite












I am trying to understand the Kronecker Delta and want to clarify. Considering the definition of the Kronecker Delta and assuming $i=j=k$ for the following situations:



I know that $delta _j^i delta _i^j $ is equal to $N$ where $N$ is the dimension of space. Would this mean that $delta _j^i delta _k^j delta _i^k $ would also be equal to $N$?



Similarly, since $delta _i^i delta _j^j $ is equal to $N^2$, would $delta _i^i delta _j^j delta _k^k $ be equal to $N^3$ ?



Last, is $delta _i^j delta _j^k $ just equal to $delta _i^k = 1 $ ?






vectors tensors kronecker-symbol kronecker-delta







share|cite|improve this question
























  • @AlvinJin This should be an answer not a comment
    – Eddy
    Sep 25 at 20:51










  • If $i=k$, wouldn't the $delta_i^k$ be equal to 1, by definition though?
    – Cave Johnson
    Sep 25 at 20:53













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am trying to understand the Kronecker Delta and want to clarify. Considering the definition of the Kronecker Delta and assuming $i=j=k$ for the following situations:



I know that $delta _j^i delta _i^j $ is equal to $N$ where $N$ is the dimension of space. Would this mean that $delta _j^i delta _k^j delta _i^k $ would also be equal to $N$?



Similarly, since $delta _i^i delta _j^j $ is equal to $N^2$, would $delta _i^i delta _j^j delta _k^k $ be equal to $N^3$ ?



Last, is $delta _i^j delta _j^k $ just equal to $delta _i^k = 1 $ ?






vectors tensors kronecker-symbol kronecker-delta







share|cite|improve this question















I am trying to understand the Kronecker Delta and want to clarify. Considering the definition of the Kronecker Delta and assuming $i=j=k$ for the following situations:



I know that $delta _j^i delta _i^j $ is equal to $N$ where $N$ is the dimension of space. Would this mean that $delta _j^i delta _k^j delta _i^k $ would also be equal to $N$?



Similarly, since $delta _i^i delta _j^j $ is equal to $N^2$, would $delta _i^i delta _j^j delta _k^k $ be equal to $N^3$ ?



Last, is $delta _i^j delta _j^k $ just equal to $delta _i^k = 1 $ ?






vectors tensors kronecker-symbol kronecker-delta




vectors tensors kronecker-symbol kronecker-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 25 at 20:41

























asked Sep 25 at 20:10









Cave Johnson

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206












  • @AlvinJin This should be an answer not a comment
    – Eddy
    Sep 25 at 20:51










  • If $i=k$, wouldn't the $delta_i^k$ be equal to 1, by definition though?
    – Cave Johnson
    Sep 25 at 20:53


















  • @AlvinJin This should be an answer not a comment
    – Eddy
    Sep 25 at 20:51










  • If $i=k$, wouldn't the $delta_i^k$ be equal to 1, by definition though?
    – Cave Johnson
    Sep 25 at 20:53
















@AlvinJin This should be an answer not a comment
– Eddy
Sep 25 at 20:51




@AlvinJin This should be an answer not a comment
– Eddy
Sep 25 at 20:51












If $i=k$, wouldn't the $delta_i^k$ be equal to 1, by definition though?
– Cave Johnson
Sep 25 at 20:53




If $i=k$, wouldn't the $delta_i^k$ be equal to 1, by definition though?
– Cave Johnson
Sep 25 at 20:53










1 Answer
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It looks to me like you're inherently using the Einstein summation convention, familiar in general relativity. For the purposes of this question, this means that any indices seen twice are summed over.



When you say $delta^i_j delta^j_i = N$, for example, this implicitly means $displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N delta^i_j delta^j_i = N$ which is indeed true since the sum is non-vanishing whenever $i=j$ and this happens $N$ times.



Similarly, $delta^i_j delta^j_k delta^k_i = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_j delta^j_k delta^k_i = N$, as you surmise.



Also $delta^i_i delta^j_j delta^k_k = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_i delta^j_j delta^k_k = N^3$, as you suspected.



Finally, $delta^j_i delta^k_j = delta^k_i$ but this is not a scalar quantity ($neq 1$). Instead it has a separate value for each index $i$ and $k$. You can think of it as being represented by a matrix whose $(i,k)$th entry is $1$ if $i=k$ and $0$ otherwise (identity matrix in $N$ dimensions).



This is a bit of a simplification of the whole picture but the essence of the machinery is here.






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    1 Answer
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    1 Answer
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    active

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    up vote
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    accepted










    It looks to me like you're inherently using the Einstein summation convention, familiar in general relativity. For the purposes of this question, this means that any indices seen twice are summed over.



    When you say $delta^i_j delta^j_i = N$, for example, this implicitly means $displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N delta^i_j delta^j_i = N$ which is indeed true since the sum is non-vanishing whenever $i=j$ and this happens $N$ times.



    Similarly, $delta^i_j delta^j_k delta^k_i = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_j delta^j_k delta^k_i = N$, as you surmise.



    Also $delta^i_i delta^j_j delta^k_k = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_i delta^j_j delta^k_k = N^3$, as you suspected.



    Finally, $delta^j_i delta^k_j = delta^k_i$ but this is not a scalar quantity ($neq 1$). Instead it has a separate value for each index $i$ and $k$. You can think of it as being represented by a matrix whose $(i,k)$th entry is $1$ if $i=k$ and $0$ otherwise (identity matrix in $N$ dimensions).



    This is a bit of a simplification of the whole picture but the essence of the machinery is here.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      It looks to me like you're inherently using the Einstein summation convention, familiar in general relativity. For the purposes of this question, this means that any indices seen twice are summed over.



      When you say $delta^i_j delta^j_i = N$, for example, this implicitly means $displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N delta^i_j delta^j_i = N$ which is indeed true since the sum is non-vanishing whenever $i=j$ and this happens $N$ times.



      Similarly, $delta^i_j delta^j_k delta^k_i = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_j delta^j_k delta^k_i = N$, as you surmise.



      Also $delta^i_i delta^j_j delta^k_k = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_i delta^j_j delta^k_k = N^3$, as you suspected.



      Finally, $delta^j_i delta^k_j = delta^k_i$ but this is not a scalar quantity ($neq 1$). Instead it has a separate value for each index $i$ and $k$. You can think of it as being represented by a matrix whose $(i,k)$th entry is $1$ if $i=k$ and $0$ otherwise (identity matrix in $N$ dimensions).



      This is a bit of a simplification of the whole picture but the essence of the machinery is here.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        It looks to me like you're inherently using the Einstein summation convention, familiar in general relativity. For the purposes of this question, this means that any indices seen twice are summed over.



        When you say $delta^i_j delta^j_i = N$, for example, this implicitly means $displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N delta^i_j delta^j_i = N$ which is indeed true since the sum is non-vanishing whenever $i=j$ and this happens $N$ times.



        Similarly, $delta^i_j delta^j_k delta^k_i = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_j delta^j_k delta^k_i = N$, as you surmise.



        Also $delta^i_i delta^j_j delta^k_k = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_i delta^j_j delta^k_k = N^3$, as you suspected.



        Finally, $delta^j_i delta^k_j = delta^k_i$ but this is not a scalar quantity ($neq 1$). Instead it has a separate value for each index $i$ and $k$. You can think of it as being represented by a matrix whose $(i,k)$th entry is $1$ if $i=k$ and $0$ otherwise (identity matrix in $N$ dimensions).



        This is a bit of a simplification of the whole picture but the essence of the machinery is here.






        share|cite|improve this answer












        It looks to me like you're inherently using the Einstein summation convention, familiar in general relativity. For the purposes of this question, this means that any indices seen twice are summed over.



        When you say $delta^i_j delta^j_i = N$, for example, this implicitly means $displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N delta^i_j delta^j_i = N$ which is indeed true since the sum is non-vanishing whenever $i=j$ and this happens $N$ times.



        Similarly, $delta^i_j delta^j_k delta^k_i = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_j delta^j_k delta^k_i = N$, as you surmise.



        Also $delta^i_i delta^j_j delta^k_k = displaystyle sum_{i=1}^N displaystyle sum_{j=1}^N displaystyle sum_{k=1}^N delta^i_i delta^j_j delta^k_k = N^3$, as you suspected.



        Finally, $delta^j_i delta^k_j = delta^k_i$ but this is not a scalar quantity ($neq 1$). Instead it has a separate value for each index $i$ and $k$. You can think of it as being represented by a matrix whose $(i,k)$th entry is $1$ if $i=k$ and $0$ otherwise (identity matrix in $N$ dimensions).



        This is a bit of a simplification of the whole picture but the essence of the machinery is here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 26 at 23:36









        hexomino

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