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Direct Sum of Alternating k-tensors

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$begingroup$

I'm reading Loring Tu's "An Introduction to Manifolds", and I'm being tripped up by a statement made in passing on p. 30:

For a finite-dimensional vector space $V$, say of dimension n, define
$$A_*(V) = bigoplus_k=0^infty A_k(V) = bigoplus_k=0^n A_k(V)$$

Here, I understand the $A_k(V)$ components to be the $k$-covectors on $V$, i.e. the set of all alternating $k$-linear functions from $V^k$ to the real numbers. What I don't understand is why the infinite direct sum of these vector spaces is equivalent to the finite direct sum truncated at $n$.

I imagine this has something to do with the dimension of $V$ itself. though I can't see why, for example, an alternating $(n+1)$-linear function would not be admissible here. I think it's likely that I'm not interpreting the notion of a direct sum of vector spaces correctly.

abstract-algebra

share|cite|improve this question

edited Mar 28 at 21:43

Captain Lama

10.1k1030

asked Mar 28 at 21:22

John GillingJohn Gilling

33

$endgroup$

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  The only alternating $k$ linear function on an $n$ dimensional vector space, with $n < k$, is zero. (Exercise.)
  $endgroup$
  – Lorenzo
  Mar 28 at 21:40
  

  
  
  
  


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  @Lorenzo thank you, that makes sense, and is actually proven quite elegantly a few pages later in the text. A follow-up question then: how is the direct sum meant to be interpreted? To be explicit, a typical element of $A_*(V)$ might look something like $a_i_0 + ... + a_i_n$, with $a_i_k in A_k(V)$, but I'm kind of unclear as to what it means to add two functions that have entirely different domains (i.e. $V$ vs $V times V$ vs. etc.).
  $endgroup$
  – John Gilling
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it's not necessarily possible to interpret the mixed tensors as functions. It's like the joke: what's an apple plus an orange? An element of the free vector space on apple, orange.
  $endgroup$
  – Lorenzo
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can add them in a formal way by extending them by zero - so they'll become functions on the disjoint union of the products of the V. So you could think of it as a function that takes in a tuple of vectors, with no restrictions on the size of that tuple, and has the appropriate multilinearity.
  $endgroup$
  – Lorenzo
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm reading Loring Tu's "An Introduction to Manifolds", and I'm being tripped up by a statement made in passing on p. 30:

For a finite-dimensional vector space $V$, say of dimension n, define
$$A_*(V) = bigoplus_k=0^infty A_k(V) = bigoplus_k=0^n A_k(V)$$

Here, I understand the $A_k(V)$ components to be the $k$-covectors on $V$, i.e. the set of all alternating $k$-linear functions from $V^k$ to the real numbers. What I don't understand is why the infinite direct sum of these vector spaces is equivalent to the finite direct sum truncated at $n$.

I imagine this has something to do with the dimension of $V$ itself. though I can't see why, for example, an alternating $(n+1)$-linear function would not be admissible here. I think it's likely that I'm not interpreting the notion of a direct sum of vector spaces correctly.

abstract-algebra

share|cite|improve this question

edited Mar 28 at 21:43

Captain Lama

10.1k1030

asked Mar 28 at 21:22

John GillingJohn Gilling

33

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The only alternating $k$ linear function on an $n$ dimensional vector space, with $n < k$, is zero. (Exercise.)
  $endgroup$
  – Lorenzo
  Mar 28 at 21:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo thank you, that makes sense, and is actually proven quite elegantly a few pages later in the text. A follow-up question then: how is the direct sum meant to be interpreted? To be explicit, a typical element of $A_*(V)$ might look something like $a_i_0 + ... + a_i_n$, with $a_i_k in A_k(V)$, but I'm kind of unclear as to what it means to add two functions that have entirely different domains (i.e. $V$ vs $V times V$ vs. etc.).
  $endgroup$
  – John Gilling
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it's not necessarily possible to interpret the mixed tensors as functions. It's like the joke: what's an apple plus an orange? An element of the free vector space on apple, orange.
  $endgroup$
  – Lorenzo
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can add them in a formal way by extending them by zero - so they'll become functions on the disjoint union of the products of the V. So you could think of it as a function that takes in a tuple of vectors, with no restrictions on the size of that tuple, and has the appropriate multilinearity.
  $endgroup$
  – Lorenzo
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm reading Loring Tu's "An Introduction to Manifolds", and I'm being tripped up by a statement made in passing on p. 30:

For a finite-dimensional vector space $V$, say of dimension n, define
$$A_*(V) = bigoplus_k=0^infty A_k(V) = bigoplus_k=0^n A_k(V)$$

Here, I understand the $A_k(V)$ components to be the $k$-covectors on $V$, i.e. the set of all alternating $k$-linear functions from $V^k$ to the real numbers. What I don't understand is why the infinite direct sum of these vector spaces is equivalent to the finite direct sum truncated at $n$.

I imagine this has something to do with the dimension of $V$ itself. though I can't see why, for example, an alternating $(n+1)$-linear function would not be admissible here. I think it's likely that I'm not interpreting the notion of a direct sum of vector spaces correctly.

abstract-algebra

share|cite|improve this question

edited Mar 28 at 21:43

Captain Lama

10.1k1030

asked Mar 28 at 21:22

John GillingJohn Gilling

33

$endgroup$

share|cite|improve this question

edited Mar 28 at 21:43

Captain Lama

10.1k1030

asked Mar 28 at 21:22

John GillingJohn Gilling

33

share|cite|improve this question

edited Mar 28 at 21:43

Captain Lama

10.1k1030

asked Mar 28 at 21:22

John GillingJohn Gilling

33

edited Mar 28 at 21:43

Captain Lama

10.1k1030

edited Mar 28 at 21:43

Captain Lama

10.1k1030

asked Mar 28 at 21:22

John GillingJohn Gilling

33

asked Mar 28 at 21:22

John GillingJohn Gilling

33