On the definition of k-simplex
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Usually, a $k$-simplex is defined as a subset of $mathbbR^n$ spanned by $k+1$ vertices $v_0,ldots , v_k$, that is the set of linear combinations of the vertices with real non negative coefficients whose sum is $1$. I have two questions: is it possible to substitute $mathbbR^n$ with any real vector space? Has that vector space to be necessarily finite dimensional over $mathbbR$?
algebraic-topology
asked Aug 14 at 7:26
bateman
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Usually, a $k$-simplex is defined as a subset of $mathbbR^n$ spanned by $k+1$ vertices $v_0,ldots , v_k$, that is the set of linear combinations of the vertices with real non negative coefficients whose sum is $1$. I have two questions: is it possible to substitute $mathbbR^n$ with any real vector space? Has that vector space to be necessarily finite dimensional over $mathbbR$?
algebraic-topology
asked Aug 14 at 7:26
bateman
1,800919
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Usually, a $k$-simplex is defined as a subset of $mathbbR^n$ spanned by $k+1$ vertices $v_0,ldots , v_k$, that is the set of linear combinations of the vertices with real non negative coefficients whose sum is $1$. I have two questions: is it possible to substitute $mathbbR^n$ with any real vector space? Has that vector space to be necessarily finite dimensional over $mathbbR$?
algebraic-topology
asked Aug 14 at 7:26
bateman
1,800919
Usually, a $k$-simplex is defined as a subset of $mathbbR^n$ spanned by $k+1$ vertices $v_0,ldots , v_k$, that is the set of linear combinations of the vertices with real non negative coefficients whose sum is $1$. I have two questions: is it possible to substitute $mathbbR^n$ with any real vector space? Has that vector space to be necessarily finite dimensional over $mathbbR$?
algebraic-topology
asked Aug 14 at 7:26
bateman
1,800919
asked Aug 14 at 7:26
bateman
1,800919
asked Aug 14 at 7:26
bateman
1,800919
asked Aug 14 at 7:26
bateman
1,800919
1,800919
add a comment |Â
add a comment |Â
1 Answer
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Nothing in the definition of simplex stops you from defining a $k$-simplplex in an infinite dimensional vector space, so the answer to your second question is 'no the vector space does not need to be finite-dimensional'.
As for the first question, as long as you are able to form finite linear combinations you can form a simplex, albeit if you find yourself dealing with a vector space without an inner product, you may have a hard time verifying whether or not the $v_0,ldots,v_k$ are in general position.
answered Aug 14 at 8:01
John C
786311
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Nothing in the definition of simplex stops you from defining a $k$-simplplex in an infinite dimensional vector space, so the answer to your second question is 'no the vector space does not need to be finite-dimensional'.
As for the first question, as long as you are able to form finite linear combinations you can form a simplex, albeit if you find yourself dealing with a vector space without an inner product, you may have a hard time verifying whether or not the $v_0,ldots,v_k$ are in general position.
answered Aug 14 at 8:01
John C
786311
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
add a comment |Â
up vote
1
down vote
accepted
Nothing in the definition of simplex stops you from defining a $k$-simplplex in an infinite dimensional vector space, so the answer to your second question is 'no the vector space does not need to be finite-dimensional'.
As for the first question, as long as you are able to form finite linear combinations you can form a simplex, albeit if you find yourself dealing with a vector space without an inner product, you may have a hard time verifying whether or not the $v_0,ldots,v_k$ are in general position.
answered Aug 14 at 8:01
John C
786311
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Nothing in the definition of simplex stops you from defining a $k$-simplplex in an infinite dimensional vector space, so the answer to your second question is 'no the vector space does not need to be finite-dimensional'.
As for the first question, as long as you are able to form finite linear combinations you can form a simplex, albeit if you find yourself dealing with a vector space without an inner product, you may have a hard time verifying whether or not the $v_0,ldots,v_k$ are in general position.
answered Aug 14 at 8:01
John C
786311
Nothing in the definition of simplex stops you from defining a $k$-simplplex in an infinite dimensional vector space, so the answer to your second question is 'no the vector space does not need to be finite-dimensional'.
As for the first question, as long as you are able to form finite linear combinations you can form a simplex, albeit if you find yourself dealing with a vector space without an inner product, you may have a hard time verifying whether or not the $v_0,ldots,v_k$ are in general position.
answered Aug 14 at 8:01
John C
786311
answered Aug 14 at 8:01
John C
786311
answered Aug 14 at 8:01
John C
786311
answered Aug 14 at 8:01
John C
786311
786311
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
add a comment |Â
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Thank you for your answer. I have a little remark: why do the inner product is necessary? Suppose there is not an inner product. Can I just say the vertices are in general position if their differences are linearly independent?
– bateman
Aug 14 at 8:06
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
Yes, that holds in any vector space, the problem I see is exactly checking whether or not they are linearly independent without the inner product. But in theory there is no problem.
– John C
Aug 14 at 12:28
add a comment |Â
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