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'tensor product of chain complexes'
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| Title of object: |
tensor product of chain complexes |
| Canonical Name: |
TensorProductOfChainComplexes |
| Type: |
Definition |
| Created on: |
2006-09-07 04:30:20 |
| Modified on: |
2006-09-07 04:31:22 |
| Classification: |
msc:16E05 |
| Keywords: |
chain complex |
| Defines: |
tensor product of chain complexes |
Preamble:
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Content:
\newcommand{\cbra}[1]{\left( #1 \right)}
\newcommand{\qbra}[1]{\left[ #1 \right]}
\newcommand{\gbra}[1]{\left\{ #1 \right\}}
\newcommand{\abra}[1]{\left\langle #1 \right\rangle}
\newcommand{\pa}[1]{\partial_{#1}}
\newcommand{\pap}[1]{\partial_{#1} '}
\newcommand{\papp}[1]{\partial_{#1} ''}
Let $C'=\gbra{C_n',\pap n}$ and $C''=\gbra{C_n'',\papp n}$ be two chain complex, their \emph{tensor product} $C'\otimes C''=\gbra{(C'\otimes C'')_n,\pa n}$ is the chain complex defined by
$$ (C'\otimes C'')_n = \bigoplus_{i+j=n}(C_i'\otimes C_j''), $$
$$ \pa n(t'_i\otimes s''_j) = \pap i(t'_i)\otimes s''_j + (-1)^i t'_i\otimes \papp j(s''_j),\ \ \ \forall t'_i\in C_i', s''_j\in C_j'',\ (i+j=n).$$
This is a good definition, in fact for each $t'_i\otimes s''_j\in C_i'\otimes C_j''\subseteq (C'\otimes C'')_{i+j}$ we have
$$\pa{i+j-1} \pa {i+j}(t'_i\otimes s''_j)=\pa{i+j-1}\cbra{ \pap i(t'_i)\otimes s''_j + (-1)^i t'_i\otimes \papp j(s''_j) }= (-1)^{i-1} \pap i(t'_i)\otimes \papp j(s''_j)+(-1)^i \pap i(t'_i)\otimes \papp j(s''_j)=0,$$
thus $C'\otimes C''$ is a chain complex. |
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