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Excluding basis in tensor notation

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0

1

$begingroup$

My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?

Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?

I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.

tensors

share|cite|improve this question

edited Mar 14 at 8:53

Benjamin Thoburn

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might find part of this answer helpful.
  $endgroup$
  – Michael Albanese
  Mar 12 at 21:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael Albanese Thanks, which part exactly?
  $endgroup$
  – Benjamin Thoburn
  Mar 14 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
  $endgroup$
  – Benjamin Thoburn
  Mar 14 at 9:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
  $endgroup$
  – Michael Albanese
  Mar 14 at 14:03
  

  
  
  
  

add a comment | 

0

1

$begingroup$

My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?

Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?

I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.

tensors

share|cite|improve this question

edited Mar 14 at 8:53

Benjamin Thoburn

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might find part of this answer helpful.
  $endgroup$
  – Michael Albanese
  Mar 12 at 21:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael Albanese Thanks, which part exactly?
  $endgroup$
  – Benjamin Thoburn
  Mar 14 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
  $endgroup$
  – Benjamin Thoburn
  Mar 14 at 9:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
  $endgroup$
  – Michael Albanese
  Mar 14 at 14:03
  

  
  
  
  

add a comment | 

$begingroup$

My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?

Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?

I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.

tensors

share|cite|improve this question

edited Mar 14 at 8:53

Benjamin Thoburn

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

$endgroup$

share|cite|improve this question

edited Mar 14 at 8:53

Benjamin Thoburn

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

share|cite|improve this question

edited Mar 14 at 8:53

Benjamin Thoburn

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

asked Mar 12 at 20:00

Benjamin ThoburnBenjamin Thoburn

356313

1 Answer
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1

$begingroup$

Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by

$mathbf T^ij mathbf e_i otimes mathbf e_j$

which is short for

$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$

But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.

share|cite|improve this answer

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825

$endgroup$

add a comment | 

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1

$begingroup$

Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by

$mathbf T^ij mathbf e_i otimes mathbf e_j$

which is short for

$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$

But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.

share|cite|improve this answer

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825

$endgroup$

add a comment | 

1

$begingroup$

Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by

$mathbf T^ij mathbf e_i otimes mathbf e_j$

which is short for

$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$

But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.

share|cite|improve this answer

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825

$endgroup$

add a comment | 

$begingroup$

Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by

$mathbf T^ij mathbf e_i otimes mathbf e_j$

which is short for

$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$

But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.

share|cite|improve this answer

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825

$endgroup$

share|cite|improve this answer

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825

answered Mar 14 at 10:06

gandalf61gandalf61

9,109825