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What “are†iterated Hom-spaces?
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Let $V$ and $W$ be vector spaces over some field $F$. Then $textrmHom(V, W)$ is the set of homomorphisms between them, which itself forms a vector space.
This definition seems clear enough to me, and I have some intuition for what vector space homomorphisms "are," but I'm always confused when "iterated-Hom" spaces appear; i.e., spaces resembling $textrmHom(textrmHom(V, W), U)$. Of course this is "just" the set of homomorphisms from one space to another, but this particular form seems to appear often enough that it must have some nice "meaning."
For instance, the second derivative $D^2 f$ of a real-valued map $f colon mathbbR^n to mathbbR^m$ can be thought of as a map from $mathbbR^n$ to $textrmHom(mathbbR^n, textrmHom(mathbbR^n, mathbbR^m))$. (The latter space is isomorphic to $mathbbR^n^2 m$, but that doesn't seem meaningful, or even useful when the vector spaces aren't finite dimensional.) Higher-order derivatives have analogous interpretations.
How should I intuitively think about such spaces?
abstract-algebra vector-spaces
asked Jul 21 at 22:39
rwbogl
716415
add a comment |Â
up vote
3
down vote
favorite
Let $V$ and $W$ be vector spaces over some field $F$. Then $textrmHom(V, W)$ is the set of homomorphisms between them, which itself forms a vector space.
This definition seems clear enough to me, and I have some intuition for what vector space homomorphisms "are," but I'm always confused when "iterated-Hom" spaces appear; i.e., spaces resembling $textrmHom(textrmHom(V, W), U)$. Of course this is "just" the set of homomorphisms from one space to another, but this particular form seems to appear often enough that it must have some nice "meaning."
For instance, the second derivative $D^2 f$ of a real-valued map $f colon mathbbR^n to mathbbR^m$ can be thought of as a map from $mathbbR^n$ to $textrmHom(mathbbR^n, textrmHom(mathbbR^n, mathbbR^m))$. (The latter space is isomorphic to $mathbbR^n^2 m$, but that doesn't seem meaningful, or even useful when the vector spaces aren't finite dimensional.) Higher-order derivatives have analogous interpretations.
How should I intuitively think about such spaces?
abstract-algebra vector-spaces
asked Jul 21 at 22:39
rwbogl
716415
1
The second section of this text discusses this very point.
– JuliusL33t
Jul 21 at 23:11
Iterated Hom-spaces are just Hom-spaces.
– Morgan Rodgers
Jul 21 at 23:24
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $V$ and $W$ be vector spaces over some field $F$. Then $textrmHom(V, W)$ is the set of homomorphisms between them, which itself forms a vector space.
This definition seems clear enough to me, and I have some intuition for what vector space homomorphisms "are," but I'm always confused when "iterated-Hom" spaces appear; i.e., spaces resembling $textrmHom(textrmHom(V, W), U)$. Of course this is "just" the set of homomorphisms from one space to another, but this particular form seems to appear often enough that it must have some nice "meaning."
For instance, the second derivative $D^2 f$ of a real-valued map $f colon mathbbR^n to mathbbR^m$ can be thought of as a map from $mathbbR^n$ to $textrmHom(mathbbR^n, textrmHom(mathbbR^n, mathbbR^m))$. (The latter space is isomorphic to $mathbbR^n^2 m$, but that doesn't seem meaningful, or even useful when the vector spaces aren't finite dimensional.) Higher-order derivatives have analogous interpretations.
How should I intuitively think about such spaces?
abstract-algebra vector-spaces
asked Jul 21 at 22:39
rwbogl
716415
Let $V$ and $W$ be vector spaces over some field $F$. Then $textrmHom(V, W)$ is the set of homomorphisms between them, which itself forms a vector space.
This definition seems clear enough to me, and I have some intuition for what vector space homomorphisms "are," but I'm always confused when "iterated-Hom" spaces appear; i.e., spaces resembling $textrmHom(textrmHom(V, W), U)$. Of course this is "just" the set of homomorphisms from one space to another, but this particular form seems to appear often enough that it must have some nice "meaning."
For instance, the second derivative $D^2 f$ of a real-valued map $f colon mathbbR^n to mathbbR^m$ can be thought of as a map from $mathbbR^n$ to $textrmHom(mathbbR^n, textrmHom(mathbbR^n, mathbbR^m))$. (The latter space is isomorphic to $mathbbR^n^2 m$, but that doesn't seem meaningful, or even useful when the vector spaces aren't finite dimensional.) Higher-order derivatives have analogous interpretations.
How should I intuitively think about such spaces?
abstract-algebra vector-spaces
asked Jul 21 at 22:39
rwbogl
716415
asked Jul 21 at 22:39
rwbogl
716415
asked Jul 21 at 22:39
rwbogl
716415
asked Jul 21 at 22:39
rwbogl
716415
716415
1
The second section of this text discusses this very point.
– JuliusL33t
Jul 21 at 23:11
Iterated Hom-spaces are just Hom-spaces.
– Morgan Rodgers
Jul 21 at 23:24
add a comment |Â
1
The second section of this text discusses this very point.
– JuliusL33t
Jul 21 at 23:11
Iterated Hom-spaces are just Hom-spaces.
– Morgan Rodgers
Jul 21 at 23:24
1
1
The second section of this text discusses this very point.
– JuliusL33t
Jul 21 at 23:11
The second section of this text discusses this very point.
– JuliusL33t
Jul 21 at 23:11
Iterated Hom-spaces are just Hom-spaces.
– Morgan Rodgers
Jul 21 at 23:24
Iterated Hom-spaces are just Hom-spaces.
– Morgan Rodgers
Jul 21 at 23:24
add a comment |Â
1 Answer
1
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up vote
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I think the main thing you have to know is that, for any commutative ring $R$, a,d any $R$-modules $M, N, P$, there are isomorphisms:$DeclareMathOperatorHomHom$
$$Hom_R(Motimes_R N,P)simeq Hom_Rbigl(M,Hom_R(N,P)bigr)$$
and
$$Hom_R(Motimes_R N,P)simeq mathscr L^2_R (M,N;P)$$
(the set of $R$-bilinear maps from $Mtimes N$ to $P$). As you know, with your notations, the second differential is a bilinear map from $mathbf R^ntimesmathbf R^ntomathbf R^m$.
answered Jul 21 at 22:59
Bernard
110k635103
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
I think the main thing you have to know is that, for any commutative ring $R$, a,d any $R$-modules $M, N, P$, there are isomorphisms:$DeclareMathOperatorHomHom$
$$Hom_R(Motimes_R N,P)simeq Hom_Rbigl(M,Hom_R(N,P)bigr)$$
and
$$Hom_R(Motimes_R N,P)simeq mathscr L^2_R (M,N;P)$$
(the set of $R$-bilinear maps from $Mtimes N$ to $P$). As you know, with your notations, the second differential is a bilinear map from $mathbf R^ntimesmathbf R^ntomathbf R^m$.
answered Jul 21 at 22:59
Bernard
110k635103
add a comment |Â
up vote
4
down vote
I think the main thing you have to know is that, for any commutative ring $R$, a,d any $R$-modules $M, N, P$, there are isomorphisms:$DeclareMathOperatorHomHom$
$$Hom_R(Motimes_R N,P)simeq Hom_Rbigl(M,Hom_R(N,P)bigr)$$
and
$$Hom_R(Motimes_R N,P)simeq mathscr L^2_R (M,N;P)$$
(the set of $R$-bilinear maps from $Mtimes N$ to $P$). As you know, with your notations, the second differential is a bilinear map from $mathbf R^ntimesmathbf R^ntomathbf R^m$.
answered Jul 21 at 22:59
Bernard
110k635103
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I think the main thing you have to know is that, for any commutative ring $R$, a,d any $R$-modules $M, N, P$, there are isomorphisms:$DeclareMathOperatorHomHom$
$$Hom_R(Motimes_R N,P)simeq Hom_Rbigl(M,Hom_R(N,P)bigr)$$
and
$$Hom_R(Motimes_R N,P)simeq mathscr L^2_R (M,N;P)$$
(the set of $R$-bilinear maps from $Mtimes N$ to $P$). As you know, with your notations, the second differential is a bilinear map from $mathbf R^ntimesmathbf R^ntomathbf R^m$.
answered Jul 21 at 22:59
Bernard
110k635103
I think the main thing you have to know is that, for any commutative ring $R$, a,d any $R$-modules $M, N, P$, there are isomorphisms:$DeclareMathOperatorHomHom$
$$Hom_R(Motimes_R N,P)simeq Hom_Rbigl(M,Hom_R(N,P)bigr)$$
and
$$Hom_R(Motimes_R N,P)simeq mathscr L^2_R (M,N;P)$$
(the set of $R$-bilinear maps from $Mtimes N$ to $P$). As you know, with your notations, the second differential is a bilinear map from $mathbf R^ntimesmathbf R^ntomathbf R^m$.
answered Jul 21 at 22:59
Bernard
110k635103
answered Jul 21 at 22:59
Bernard
110k635103
answered Jul 21 at 22:59
Bernard
110k635103
answered Jul 21 at 22:59
Bernard
110k635103
110k635103
add a comment |Â
add a comment |Â
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1
The second section of this text discusses this very point.
– JuliusL33t
Jul 21 at 23:11
Iterated Hom-spaces are just Hom-spaces.
– Morgan Rodgers
Jul 21 at 23:24