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Operator space tensor products
An operator space tensor product is the completion of the algebraic tensor
product with respect to an operator space tensor norm.
An
operator space tensor norm
is defined for each pair
of operator spaces and endows
their algebraic tensor product
with the structure of an
matrix normed space
thuch that the following two properties
and
[BP91, Def. 5.9].
hold.
The completion is called the
-operator space tensor product
of
and
and is denoted by
.
- 1.
- For the complex numbers holds
- 2.
-
For all
and
the operator
has a continuous extension
The bilinear mapping
is
jointly completely contractive .30
- Property
may be replaced by the assumptions
and
.31
- 3.
- An operator space tensor product
is functorial:
For all
and
the operator
has a continuous extension
and
- Remark: This is indeed an equality
.
- 4.
- The algebraic
shuffle isomorphism
has a continuous extension to a complete contraction:
This complete contraction is called the
shuffle map of the
-operator space tensor product.
Condition
is equivalent to the following two
conditions: The
shuffle mappings
are completely contractive.
Operator space tensor producte
may have further special properties:
- An operator space tensor product
is called
-
symmetric,
if
is a complete isometry;
-
associative,
if
is acomplete isometry;
-
injective,
if for all subspaces
,
the map
is acomplete isometry;
-
projective,
if for all subspaces
,
the map
is a complete quotient map;
-
self dual,
if the algebraic embedding
has a completely isometric extension
.
In many applications one finds the
Haagerup -tensor product.
It is not symmetric, but associative, injective,
projective and self dual.
Footnotes
- ...
30
- i.e., let
,
,
,
then the norm of the linear operator
is estimated by
![$\displaystyle \Vert [S_{ij} \otimes T_{kl}] \Vert _\mathrm{cb}
\leq
\Vert [S_{ij}] \Vert _\mathrm{cb}\Vert [T_{kl}] \Vert _\mathrm{cb}$](img652.png)
.
Remark: This is indeed an equality.
- ...itemnr:ORTPaxiom4.31
- (2)
(3),(4):
Condition
is a special case of
.
Let
,
be matrices, which are algebraically the the identical mappings
of the vector spaces
respectivly
.
By assumption
we have
Now
is the shuffle-map in
.
(3),(4)
(2):
Let
,
,
,
and
,
the corresponding operators.
By
holds
We apply the shuffle map
and obtain from
Hence
is jointly completely bounded.
Subsections
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Prof. Gerd Wittstock
2001-01-07