Mixing
Why are $(1,0)$ and $(0,1)$ tensors antisymmetric?
$begingroup$
The book I'm reading (Nadir Jeevanjee (auth.)-An Introduction to Tensors and Group Theory for Physicists-Birkhäuser Basel (2015)) defined an antisymmetric tensor of type $(r,0)$ or $(0,r)$ as
"one whose value changes sign under transposition of any two of its
arguments."
OK, I totally understand the definition, but now the book is telling me that all tensors of type $(1,0)$, and $(0,1)$, are antisymmetric (the Wikipedia article about antisymmetric tensors agrees and even says they are trivial examples of an antisymmetric tensor). But I just don't get it! First, in the definition the book says "under transposition of any two of its arguments", but if the tensor is of type $(1,0)$, how can I transpose two arguments when I can just plug one? The only way, in my conception, for a tensor of type $(1,0)$ or $(0,1)$ to be antisymmetric is only if it is the $0$ tensor, since $0 = -0$.
What am I getting wrong?
linear-algebra multilinear-algebra
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
$endgroup$
add a comment |
$begingroup$
The book I'm reading (Nadir Jeevanjee (auth.)-An Introduction to Tensors and Group Theory for Physicists-Birkhäuser Basel (2015)) defined an antisymmetric tensor of type $(r,0)$ or $(0,r)$ as
"one whose value changes sign under transposition of any two of its
arguments."
OK, I totally understand the definition, but now the book is telling me that all tensors of type $(1,0)$, and $(0,1)$, are antisymmetric (the Wikipedia article about antisymmetric tensors agrees and even says they are trivial examples of an antisymmetric tensor). But I just don't get it! First, in the definition the book says "under transposition of any two of its arguments", but if the tensor is of type $(1,0)$, how can I transpose two arguments when I can just plug one? The only way, in my conception, for a tensor of type $(1,0)$ or $(0,1)$ to be antisymmetric is only if it is the $0$ tensor, since $0 = -0$.
What am I getting wrong?
linear-algebra multilinear-algebra
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
$endgroup$
1
$begingroup$
Vectors and 1-forms are vaccously (vaccous truth) skew. I mean, they only have one index, so you can't transpose any two of its arguments. In this case, the assertion comes true. My explanation is really bad but you can read about that here and here. The idea is as follows: $T$ antysymmetric $Leftrightarrow (forall mbox{ indices of } T; i,j;; T_{dots idots jdots}=-T_{dots jdots idots})$ and since vectors only have 1 index the statement inside the brackes comes true..
$endgroup$
– Dog_69
Jan 17 at 21:48
$begingroup$
I understand your point, but I still consider it very strange for them to be considered antisymmetric.
$endgroup$
– Tandeitnik
Jan 17 at 22:00
$begingroup$
Another answer is: because you need it. But I don't like it too much, even it is true. It is convinitent for you to have your vector space inside the algebra.
$endgroup$
– Dog_69
Jan 17 at 22:03
add a comment |
$begingroup$
The book I'm reading (Nadir Jeevanjee (auth.)-An Introduction to Tensors and Group Theory for Physicists-Birkhäuser Basel (2015)) defined an antisymmetric tensor of type $(r,0)$ or $(0,r)$ as
"one whose value changes sign under transposition of any two of its
arguments."
OK, I totally understand the definition, but now the book is telling me that all tensors of type $(1,0)$, and $(0,1)$, are antisymmetric (the Wikipedia article about antisymmetric tensors agrees and even says they are trivial examples of an antisymmetric tensor). But I just don't get it! First, in the definition the book says "under transposition of any two of its arguments", but if the tensor is of type $(1,0)$, how can I transpose two arguments when I can just plug one? The only way, in my conception, for a tensor of type $(1,0)$ or $(0,1)$ to be antisymmetric is only if it is the $0$ tensor, since $0 = -0$.
What am I getting wrong?
linear-algebra multilinear-algebra
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
$endgroup$
The book I'm reading (Nadir Jeevanjee (auth.)-An Introduction to Tensors and Group Theory for Physicists-Birkhäuser Basel (2015)) defined an antisymmetric tensor of type $(r,0)$ or $(0,r)$ as
"one whose value changes sign under transposition of any two of its
arguments."
OK, I totally understand the definition, but now the book is telling me that all tensors of type $(1,0)$, and $(0,1)$, are antisymmetric (the Wikipedia article about antisymmetric tensors agrees and even says they are trivial examples of an antisymmetric tensor). But I just don't get it! First, in the definition the book says "under transposition of any two of its arguments", but if the tensor is of type $(1,0)$, how can I transpose two arguments when I can just plug one? The only way, in my conception, for a tensor of type $(1,0)$ or $(0,1)$ to be antisymmetric is only if it is the $0$ tensor, since $0 = -0$.
What am I getting wrong?
linear-algebra multilinear-algebra
linear-algebra multilinear-algebra
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
edited Jan 17 at 22:09
J. W. Tanner
2,5081117
2,5081117
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
asked Jan 17 at 21:36
TandeitnikTandeitnik
233
233
1
$begingroup$
Vectors and 1-forms are vaccously (vaccous truth) skew. I mean, they only have one index, so you can't transpose any two of its arguments. In this case, the assertion comes true. My explanation is really bad but you can read about that here and here. The idea is as follows: $T$ antysymmetric $Leftrightarrow (forall mbox{ indices of } T; i,j;; T_{dots idots jdots}=-T_{dots jdots idots})$ and since vectors only have 1 index the statement inside the brackes comes true..
$endgroup$
– Dog_69
Jan 17 at 21:48
$begingroup$
I understand your point, but I still consider it very strange for them to be considered antisymmetric.
$endgroup$
– Tandeitnik
Jan 17 at 22:00
$begingroup$
Another answer is: because you need it. But I don't like it too much, even it is true. It is convinitent for you to have your vector space inside the algebra.
$endgroup$
– Dog_69
Jan 17 at 22:03
add a comment |
1
$begingroup$
Vectors and 1-forms are vaccously (vaccous truth) skew. I mean, they only have one index, so you can't transpose any two of its arguments. In this case, the assertion comes true. My explanation is really bad but you can read about that here and here. The idea is as follows: $T$ antysymmetric $Leftrightarrow (forall mbox{ indices of } T; i,j;; T_{dots idots jdots}=-T_{dots jdots idots})$ and since vectors only have 1 index the statement inside the brackes comes true..
$endgroup$
– Dog_69
Jan 17 at 21:48
$begingroup$
I understand your point, but I still consider it very strange for them to be considered antisymmetric.
$endgroup$
– Tandeitnik
Jan 17 at 22:00
$begingroup$
Another answer is: because you need it. But I don't like it too much, even it is true. It is convinitent for you to have your vector space inside the algebra.
$endgroup$
– Dog_69
Jan 17 at 22:03
1
1
$begingroup$
Vectors and 1-forms are vaccously (vaccous truth) skew. I mean, they only have one index, so you can't transpose any two of its arguments. In this case, the assertion comes true. My explanation is really bad but you can read about that here and here. The idea is as follows: $T$ antysymmetric $Leftrightarrow (forall mbox{ indices of } T; i,j;; T_{dots idots jdots}=-T_{dots jdots idots})$ and since vectors only have 1 index the statement inside the brackes comes true..
$endgroup$
– Dog_69
Jan 17 at 21:48
$begingroup$
Vectors and 1-forms are vaccously (vaccous truth) skew. I mean, they only have one index, so you can't transpose any two of its arguments. In this case, the assertion comes true. My explanation is really bad but you can read about that here and here. The idea is as follows: $T$ antysymmetric $Leftrightarrow (forall mbox{ indices of } T; i,j;; T_{dots idots jdots}=-T_{dots jdots idots})$ and since vectors only have 1 index the statement inside the brackes comes true..
$endgroup$
– Dog_69
Jan 17 at 21:48
$begingroup$
I understand your point, but I still consider it very strange for them to be considered antisymmetric.
$endgroup$
– Tandeitnik
Jan 17 at 22:00
$begingroup$
I understand your point, but I still consider it very strange for them to be considered antisymmetric.
$endgroup$
– Tandeitnik
Jan 17 at 22:00
$begingroup$
Another answer is: because you need it. But I don't like it too much, even it is true. It is convinitent for you to have your vector space inside the algebra.
$endgroup$
– Dog_69
Jan 17 at 22:03
$begingroup$
Another answer is: because you need it. But I don't like it too much, even it is true. It is convinitent for you to have your vector space inside the algebra.
$endgroup$
– Dog_69
Jan 17 at 22:03
add a comment |
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$begingroup$
Vectors and 1-forms are vaccously (vaccous truth) skew. I mean, they only have one index, so you can't transpose any two of its arguments. In this case, the assertion comes true. My explanation is really bad but you can read about that here and here. The idea is as follows: $T$ antysymmetric $Leftrightarrow (forall mbox{ indices of } T; i,j;; T_{dots idots jdots}=-T_{dots jdots idots})$ and since vectors only have 1 index the statement inside the brackes comes true..
$endgroup$
– Dog_69
Jan 17 at 21:48
$begingroup$
I understand your point, but I still consider it very strange for them to be considered antisymmetric.
$endgroup$
– Tandeitnik
Jan 17 at 22:00
$begingroup$
Another answer is: because you need it. But I don't like it too much, even it is true. It is convinitent for you to have your vector space inside the algebra.
$endgroup$
– Dog_69
Jan 17 at 22:03