What do “projections” out of tensor products look like?
$begingroup$
Convention. If $R$ is a relation $X rightarrow Y$, what I mean is that $R$ is a subsets of $X times Y$. We say that $X$ is the domain of $R$ and that $Y$ is the codomain. We'll write $x overset{R}{mapsto} y$ to mean $(x,y) in R$. Define also that $R(x) = {y in Y : (x,y) in R}.$
Let $X$ and $Y$ denote $mathbb{R}$-modules. Then there is, in general, no meaningful choice of projection map $X otimes Y rightarrow X$. There does, however, exist a perfectly good relation $$R : X otimes Y rightarrow X$$
defined as follows: it's the smallest relation with the above domain and codomain subject to the constraint that for all finite sets $I$ and all $I$-indexed sequences
- $x : I rightarrow X$
- $y : I rightarrow Y,$
we have:
$$sum_{i in I}x_iotimes y_i overset{R}{mapsto} sum_{i in I} x_i.$$
In general $R$ will not be deterministic. In particular, note that
$$sum_{i in I}a_ix_iotimes y_i = sum_{i in I}x_iotimes a_iy_i overset{R}{mapsto} sum_{i in I} x_i.$$
This means that we can scale the $x_i$'s by non-zero elements without impacting the output of $R$.
Question. Is there a good description of the subset of $X$ given by the following expression? $$Rleft(sum_{i in I}a_ix_iotimes y_iright)$$
vector-spaces modules linear-transformations tensor-products
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
$endgroup$
add a comment |
$begingroup$
Convention. If $R$ is a relation $X rightarrow Y$, what I mean is that $R$ is a subsets of $X times Y$. We say that $X$ is the domain of $R$ and that $Y$ is the codomain. We'll write $x overset{R}{mapsto} y$ to mean $(x,y) in R$. Define also that $R(x) = {y in Y : (x,y) in R}.$
Let $X$ and $Y$ denote $mathbb{R}$-modules. Then there is, in general, no meaningful choice of projection map $X otimes Y rightarrow X$. There does, however, exist a perfectly good relation $$R : X otimes Y rightarrow X$$
defined as follows: it's the smallest relation with the above domain and codomain subject to the constraint that for all finite sets $I$ and all $I$-indexed sequences
- $x : I rightarrow X$
- $y : I rightarrow Y,$
we have:
$$sum_{i in I}x_iotimes y_i overset{R}{mapsto} sum_{i in I} x_i.$$
In general $R$ will not be deterministic. In particular, note that
$$sum_{i in I}a_ix_iotimes y_i = sum_{i in I}x_iotimes a_iy_i overset{R}{mapsto} sum_{i in I} x_i.$$
This means that we can scale the $x_i$'s by non-zero elements without impacting the output of $R$.
Question. Is there a good description of the subset of $X$ given by the following expression? $$Rleft(sum_{i in I}a_ix_iotimes y_iright)$$
vector-spaces modules linear-transformations tensor-products
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
$endgroup$
add a comment |
$begingroup$
Convention. If $R$ is a relation $X rightarrow Y$, what I mean is that $R$ is a subsets of $X times Y$. We say that $X$ is the domain of $R$ and that $Y$ is the codomain. We'll write $x overset{R}{mapsto} y$ to mean $(x,y) in R$. Define also that $R(x) = {y in Y : (x,y) in R}.$
Let $X$ and $Y$ denote $mathbb{R}$-modules. Then there is, in general, no meaningful choice of projection map $X otimes Y rightarrow X$. There does, however, exist a perfectly good relation $$R : X otimes Y rightarrow X$$
defined as follows: it's the smallest relation with the above domain and codomain subject to the constraint that for all finite sets $I$ and all $I$-indexed sequences
- $x : I rightarrow X$
- $y : I rightarrow Y,$
we have:
$$sum_{i in I}x_iotimes y_i overset{R}{mapsto} sum_{i in I} x_i.$$
In general $R$ will not be deterministic. In particular, note that
$$sum_{i in I}a_ix_iotimes y_i = sum_{i in I}x_iotimes a_iy_i overset{R}{mapsto} sum_{i in I} x_i.$$
This means that we can scale the $x_i$'s by non-zero elements without impacting the output of $R$.
Question. Is there a good description of the subset of $X$ given by the following expression? $$Rleft(sum_{i in I}a_ix_iotimes y_iright)$$
vector-spaces modules linear-transformations tensor-products
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
$endgroup$
Convention. If $R$ is a relation $X rightarrow Y$, what I mean is that $R$ is a subsets of $X times Y$. We say that $X$ is the domain of $R$ and that $Y$ is the codomain. We'll write $x overset{R}{mapsto} y$ to mean $(x,y) in R$. Define also that $R(x) = {y in Y : (x,y) in R}.$
Let $X$ and $Y$ denote $mathbb{R}$-modules. Then there is, in general, no meaningful choice of projection map $X otimes Y rightarrow X$. There does, however, exist a perfectly good relation $$R : X otimes Y rightarrow X$$
defined as follows: it's the smallest relation with the above domain and codomain subject to the constraint that for all finite sets $I$ and all $I$-indexed sequences
- $x : I rightarrow X$
- $y : I rightarrow Y,$
we have:
$$sum_{i in I}x_iotimes y_i overset{R}{mapsto} sum_{i in I} x_i.$$
In general $R$ will not be deterministic. In particular, note that
$$sum_{i in I}a_ix_iotimes y_i = sum_{i in I}x_iotimes a_iy_i overset{R}{mapsto} sum_{i in I} x_i.$$
This means that we can scale the $x_i$'s by non-zero elements without impacting the output of $R$.
Question. Is there a good description of the subset of $X$ given by the following expression? $$Rleft(sum_{i in I}a_ix_iotimes y_iright)$$
vector-spaces modules linear-transformations tensor-products
vector-spaces modules linear-transformations tensor-products
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
asked Dec 5 '18 at 23:25
goblingoblin
36.7k1159191
36.7k1159191
add a comment |
add a comment |
1 Answer
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$begingroup$
$R$ is in fact the total relation: every element of $Xotimes Y$ is related to every element of $X$. Indeed, note that for any $xin X$, $R$ turns the expression $xotimes 0$ into $x$. So, given any element $tin Xotimes Y$, we can add a term $xotimes 0$ to our representation of $t$ to add an arbitrary $x$ to the image of $t$ under $R$, and thus $R$ can map $t$ to every single element of $X$.
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$R$ is in fact the total relation: every element of $Xotimes Y$ is related to every element of $X$. Indeed, note that for any $xin X$, $R$ turns the expression $xotimes 0$ into $x$. So, given any element $tin Xotimes Y$, we can add a term $xotimes 0$ to our representation of $t$ to add an arbitrary $x$ to the image of $t$ under $R$, and thus $R$ can map $t$ to every single element of $X$.
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
$begingroup$
$R$ is in fact the total relation: every element of $Xotimes Y$ is related to every element of $X$. Indeed, note that for any $xin X$, $R$ turns the expression $xotimes 0$ into $x$. So, given any element $tin Xotimes Y$, we can add a term $xotimes 0$ to our representation of $t$ to add an arbitrary $x$ to the image of $t$ under $R$, and thus $R$ can map $t$ to every single element of $X$.
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
$endgroup$
add a comment |
$begingroup$
$R$ is in fact the total relation: every element of $Xotimes Y$ is related to every element of $X$. Indeed, note that for any $xin X$, $R$ turns the expression $xotimes 0$ into $x$. So, given any element $tin Xotimes Y$, we can add a term $xotimes 0$ to our representation of $t$ to add an arbitrary $x$ to the image of $t$ under $R$, and thus $R$ can map $t$ to every single element of $X$.
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
$endgroup$
$R$ is in fact the total relation: every element of $Xotimes Y$ is related to every element of $X$. Indeed, note that for any $xin X$, $R$ turns the expression $xotimes 0$ into $x$. So, given any element $tin Xotimes Y$, we can add a term $xotimes 0$ to our representation of $t$ to add an arbitrary $x$ to the image of $t$ under $R$, and thus $R$ can map $t$ to every single element of $X$.
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
answered Dec 6 '18 at 0:04
Eric WofseyEric Wofsey
181k12208336
181k12208336
add a comment |
add a comment |
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