Endofunctors of $textsf{FinVec}$ which are not $C^n$
Let $textsf{FinVec}$ denote the category of finite-dimensional real vector spaces. For each vector space $V in textsf{FinVec}$, choose a vector space isomorphism $varphi colon V to mathbb{R}^{dim V}$ and endow $V$ with the standard smooth structure from $mathbb{R}^{dim V}$ pulled back along $varphi$. I think (correct me if I'm wrong!) that this smooth structure on $V$ is independent of the choice of $varphi$ up to diffeomorphism.
Definition Let $n in mathbb{N} cup {infty}$ (where we use the convention that $mathbb{N}$ includes $0$). A functor $F colon textsf{FinVec} to textsf{FinVec}$ is called $C^n$ iff
$$f mapsto F(f) colon operatorname{Hom}(V,W) to operatorname{Hom}(F(V), F(W))$$
is a $C^n$ function for all $V,W in textsf{FinVec}$ (where the smooth structures on the hom sets are the ones described above, as these hom sets are finite dimensional real vector spaces under pointwise addition & scalar multiplication).
Of course, any $C^n$ functor is also $C^m$ whenever $m leq n$, and in particular $C^infty$ functors are $C^n$ for all $n$. Examples of $C^infty$ functors include:
$operatorname{Hom}(V,{-})$ for any $V in textsf{FinVec}$
$V otimes {-}$ for any $V in textsf{FinVec}$
- The exterior algebra functor
However, I've been unable to think of any examples of an endofunctor of $textsf{FinVec}$ which is not $C^infty$! So:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^infty$?
If the answer is yes, then my follow-up questions are:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^0$? If $n,m in mathbb{N} cup {infty}$ with $m < n$, does there exist an endofunctor of $textsf{FinVec}$ which is $C^m$ but not $C^n$?
linear-algebra smooth-manifolds
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
|
show 4 more comments
Let $textsf{FinVec}$ denote the category of finite-dimensional real vector spaces. For each vector space $V in textsf{FinVec}$, choose a vector space isomorphism $varphi colon V to mathbb{R}^{dim V}$ and endow $V$ with the standard smooth structure from $mathbb{R}^{dim V}$ pulled back along $varphi$. I think (correct me if I'm wrong!) that this smooth structure on $V$ is independent of the choice of $varphi$ up to diffeomorphism.
Definition Let $n in mathbb{N} cup {infty}$ (where we use the convention that $mathbb{N}$ includes $0$). A functor $F colon textsf{FinVec} to textsf{FinVec}$ is called $C^n$ iff
$$f mapsto F(f) colon operatorname{Hom}(V,W) to operatorname{Hom}(F(V), F(W))$$
is a $C^n$ function for all $V,W in textsf{FinVec}$ (where the smooth structures on the hom sets are the ones described above, as these hom sets are finite dimensional real vector spaces under pointwise addition & scalar multiplication).
Of course, any $C^n$ functor is also $C^m$ whenever $m leq n$, and in particular $C^infty$ functors are $C^n$ for all $n$. Examples of $C^infty$ functors include:
$operatorname{Hom}(V,{-})$ for any $V in textsf{FinVec}$
$V otimes {-}$ for any $V in textsf{FinVec}$
- The exterior algebra functor
However, I've been unable to think of any examples of an endofunctor of $textsf{FinVec}$ which is not $C^infty$! So:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^infty$?
If the answer is yes, then my follow-up questions are:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^0$? If $n,m in mathbb{N} cup {infty}$ with $m < n$, does there exist an endofunctor of $textsf{FinVec}$ which is $C^m$ but not $C^n$?
linear-algebra smooth-manifolds
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
2
To clarify, over $mathbb{C}$ there are "wild" endofunctors coming from composing scalar multiplication with discontinuous endomorphisms $mathbb{C} to mathbb{C}$, of which there are many (because $mathbb{C}$ is algebraically closed). But $mathbb{R}$ has no nontrivial endomorphisms as a ring. All the endofunctors I know over $mathbb{R}$ are direct sums of Schur functors, which are $C^{infty}$ and even algebraic: en.wikipedia.org/wiki/Schur_functor
– Qiaochu Yuan
Nov 28 '18 at 4:17
1
@QiaochuYuan To check my understanding: take a discontinuous ring automorphism $alpha$ of $mathbb{C}$. For each $mathbb{C}$-vector space $V$, fix a basis and let $overline{alpha}_V : V to V$ be the function $(x_1,dots,x_n) mapsto (alpha(x_1),dots,alpha(x_n))$. Now we can construct an endofunctor $F$ of $textsf{FinVec}_mathbb{C}$ by sending each object to itself and a linear map $f : V to W$ to $overline{alpha}_W circ f circ overline{alpha}_V^{-1}$. This functor is not $C^0$ because (e.g.) the map on $text{Hom}(mathbb{C},mathbb{C})$ is canonically diffeo. to $alpha$.
– diracdeltafunk
Nov 28 '18 at 5:40
1
@diracdeltafunk: it's unnecessary to fix a basis, and not quite. A vector space over $mathbb{C}$ is an abelian group $A$ together with an action $rho : mathbb{C} to text{End}(A)$. If $alpha : mathbb{C} to mathbb{C}$ is an endomorphism, then given an action $rho$ you can produce a new action $rho circ alpha$; this is the functor. Note that this construction doesn't require $alpha$ to be invertible.
– Qiaochu Yuan
Nov 28 '18 at 5:43
1
@QiaochuYuan I'm familiar with the idea of constructing a functor $f^* : R-textsf{Mod} to S-textsf{Mod}$ from a ring homomorphism $f : S to R$. However, doesn't this functor send a linear map to itself (the same function of underlying sets)? If so, won't it give the identity function $text{Hom}(V,W) to text{Hom}(V,W)$ (which is necessarily continuous)?
– diracdeltafunk
Nov 28 '18 at 5:50
3
Ooh, here's an endofunctor for the real case: compose any one of @QiaochuYuan 's wild endofunctors with the complexification functor (and then promptly forget the complex structure, of course)
– Monstrous Moonshiner
Nov 28 '18 at 6:44
|
show 4 more comments
Let $textsf{FinVec}$ denote the category of finite-dimensional real vector spaces. For each vector space $V in textsf{FinVec}$, choose a vector space isomorphism $varphi colon V to mathbb{R}^{dim V}$ and endow $V$ with the standard smooth structure from $mathbb{R}^{dim V}$ pulled back along $varphi$. I think (correct me if I'm wrong!) that this smooth structure on $V$ is independent of the choice of $varphi$ up to diffeomorphism.
Definition Let $n in mathbb{N} cup {infty}$ (where we use the convention that $mathbb{N}$ includes $0$). A functor $F colon textsf{FinVec} to textsf{FinVec}$ is called $C^n$ iff
$$f mapsto F(f) colon operatorname{Hom}(V,W) to operatorname{Hom}(F(V), F(W))$$
is a $C^n$ function for all $V,W in textsf{FinVec}$ (where the smooth structures on the hom sets are the ones described above, as these hom sets are finite dimensional real vector spaces under pointwise addition & scalar multiplication).
Of course, any $C^n$ functor is also $C^m$ whenever $m leq n$, and in particular $C^infty$ functors are $C^n$ for all $n$. Examples of $C^infty$ functors include:
$operatorname{Hom}(V,{-})$ for any $V in textsf{FinVec}$
$V otimes {-}$ for any $V in textsf{FinVec}$
- The exterior algebra functor
However, I've been unable to think of any examples of an endofunctor of $textsf{FinVec}$ which is not $C^infty$! So:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^infty$?
If the answer is yes, then my follow-up questions are:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^0$? If $n,m in mathbb{N} cup {infty}$ with $m < n$, does there exist an endofunctor of $textsf{FinVec}$ which is $C^m$ but not $C^n$?
linear-algebra smooth-manifolds
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
Let $textsf{FinVec}$ denote the category of finite-dimensional real vector spaces. For each vector space $V in textsf{FinVec}$, choose a vector space isomorphism $varphi colon V to mathbb{R}^{dim V}$ and endow $V$ with the standard smooth structure from $mathbb{R}^{dim V}$ pulled back along $varphi$. I think (correct me if I'm wrong!) that this smooth structure on $V$ is independent of the choice of $varphi$ up to diffeomorphism.
Definition Let $n in mathbb{N} cup {infty}$ (where we use the convention that $mathbb{N}$ includes $0$). A functor $F colon textsf{FinVec} to textsf{FinVec}$ is called $C^n$ iff
$$f mapsto F(f) colon operatorname{Hom}(V,W) to operatorname{Hom}(F(V), F(W))$$
is a $C^n$ function for all $V,W in textsf{FinVec}$ (where the smooth structures on the hom sets are the ones described above, as these hom sets are finite dimensional real vector spaces under pointwise addition & scalar multiplication).
Of course, any $C^n$ functor is also $C^m$ whenever $m leq n$, and in particular $C^infty$ functors are $C^n$ for all $n$. Examples of $C^infty$ functors include:
$operatorname{Hom}(V,{-})$ for any $V in textsf{FinVec}$
$V otimes {-}$ for any $V in textsf{FinVec}$
- The exterior algebra functor
However, I've been unable to think of any examples of an endofunctor of $textsf{FinVec}$ which is not $C^infty$! So:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^infty$?
If the answer is yes, then my follow-up questions are:
Does there exist an endofunctor of $textsf{FinVec}$ which is not $C^0$? If $n,m in mathbb{N} cup {infty}$ with $m < n$, does there exist an endofunctor of $textsf{FinVec}$ which is $C^m$ but not $C^n$?
linear-algebra smooth-manifolds
linear-algebra smooth-manifolds
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
edited Nov 28 '18 at 5:08
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
asked Nov 28 '18 at 3:37
diracdeltafunk
340111
340111
2
To clarify, over $mathbb{C}$ there are "wild" endofunctors coming from composing scalar multiplication with discontinuous endomorphisms $mathbb{C} to mathbb{C}$, of which there are many (because $mathbb{C}$ is algebraically closed). But $mathbb{R}$ has no nontrivial endomorphisms as a ring. All the endofunctors I know over $mathbb{R}$ are direct sums of Schur functors, which are $C^{infty}$ and even algebraic: en.wikipedia.org/wiki/Schur_functor
– Qiaochu Yuan
Nov 28 '18 at 4:17
1
@QiaochuYuan To check my understanding: take a discontinuous ring automorphism $alpha$ of $mathbb{C}$. For each $mathbb{C}$-vector space $V$, fix a basis and let $overline{alpha}_V : V to V$ be the function $(x_1,dots,x_n) mapsto (alpha(x_1),dots,alpha(x_n))$. Now we can construct an endofunctor $F$ of $textsf{FinVec}_mathbb{C}$ by sending each object to itself and a linear map $f : V to W$ to $overline{alpha}_W circ f circ overline{alpha}_V^{-1}$. This functor is not $C^0$ because (e.g.) the map on $text{Hom}(mathbb{C},mathbb{C})$ is canonically diffeo. to $alpha$.
– diracdeltafunk
Nov 28 '18 at 5:40
1
@diracdeltafunk: it's unnecessary to fix a basis, and not quite. A vector space over $mathbb{C}$ is an abelian group $A$ together with an action $rho : mathbb{C} to text{End}(A)$. If $alpha : mathbb{C} to mathbb{C}$ is an endomorphism, then given an action $rho$ you can produce a new action $rho circ alpha$; this is the functor. Note that this construction doesn't require $alpha$ to be invertible.
– Qiaochu Yuan
Nov 28 '18 at 5:43
1
@QiaochuYuan I'm familiar with the idea of constructing a functor $f^* : R-textsf{Mod} to S-textsf{Mod}$ from a ring homomorphism $f : S to R$. However, doesn't this functor send a linear map to itself (the same function of underlying sets)? If so, won't it give the identity function $text{Hom}(V,W) to text{Hom}(V,W)$ (which is necessarily continuous)?
– diracdeltafunk
Nov 28 '18 at 5:50
3
Ooh, here's an endofunctor for the real case: compose any one of @QiaochuYuan 's wild endofunctors with the complexification functor (and then promptly forget the complex structure, of course)
– Monstrous Moonshiner
Nov 28 '18 at 6:44
|
show 4 more comments
2
To clarify, over $mathbb{C}$ there are "wild" endofunctors coming from composing scalar multiplication with discontinuous endomorphisms $mathbb{C} to mathbb{C}$, of which there are many (because $mathbb{C}$ is algebraically closed). But $mathbb{R}$ has no nontrivial endomorphisms as a ring. All the endofunctors I know over $mathbb{R}$ are direct sums of Schur functors, which are $C^{infty}$ and even algebraic: en.wikipedia.org/wiki/Schur_functor
– Qiaochu Yuan
Nov 28 '18 at 4:17
1
@QiaochuYuan To check my understanding: take a discontinuous ring automorphism $alpha$ of $mathbb{C}$. For each $mathbb{C}$-vector space $V$, fix a basis and let $overline{alpha}_V : V to V$ be the function $(x_1,dots,x_n) mapsto (alpha(x_1),dots,alpha(x_n))$. Now we can construct an endofunctor $F$ of $textsf{FinVec}_mathbb{C}$ by sending each object to itself and a linear map $f : V to W$ to $overline{alpha}_W circ f circ overline{alpha}_V^{-1}$. This functor is not $C^0$ because (e.g.) the map on $text{Hom}(mathbb{C},mathbb{C})$ is canonically diffeo. to $alpha$.
– diracdeltafunk
Nov 28 '18 at 5:40
1
@diracdeltafunk: it's unnecessary to fix a basis, and not quite. A vector space over $mathbb{C}$ is an abelian group $A$ together with an action $rho : mathbb{C} to text{End}(A)$. If $alpha : mathbb{C} to mathbb{C}$ is an endomorphism, then given an action $rho$ you can produce a new action $rho circ alpha$; this is the functor. Note that this construction doesn't require $alpha$ to be invertible.
– Qiaochu Yuan
Nov 28 '18 at 5:43
1
@QiaochuYuan I'm familiar with the idea of constructing a functor $f^* : R-textsf{Mod} to S-textsf{Mod}$ from a ring homomorphism $f : S to R$. However, doesn't this functor send a linear map to itself (the same function of underlying sets)? If so, won't it give the identity function $text{Hom}(V,W) to text{Hom}(V,W)$ (which is necessarily continuous)?
– diracdeltafunk
Nov 28 '18 at 5:50
3
Ooh, here's an endofunctor for the real case: compose any one of @QiaochuYuan 's wild endofunctors with the complexification functor (and then promptly forget the complex structure, of course)
– Monstrous Moonshiner
Nov 28 '18 at 6:44
2
2
To clarify, over $mathbb{C}$ there are "wild" endofunctors coming from composing scalar multiplication with discontinuous endomorphisms $mathbb{C} to mathbb{C}$, of which there are many (because $mathbb{C}$ is algebraically closed). But $mathbb{R}$ has no nontrivial endomorphisms as a ring. All the endofunctors I know over $mathbb{R}$ are direct sums of Schur functors, which are $C^{infty}$ and even algebraic: en.wikipedia.org/wiki/Schur_functor
– Qiaochu Yuan
Nov 28 '18 at 4:17
To clarify, over $mathbb{C}$ there are "wild" endofunctors coming from composing scalar multiplication with discontinuous endomorphisms $mathbb{C} to mathbb{C}$, of which there are many (because $mathbb{C}$ is algebraically closed). But $mathbb{R}$ has no nontrivial endomorphisms as a ring. All the endofunctors I know over $mathbb{R}$ are direct sums of Schur functors, which are $C^{infty}$ and even algebraic: en.wikipedia.org/wiki/Schur_functor
– Qiaochu Yuan
Nov 28 '18 at 4:17
1
1
@QiaochuYuan To check my understanding: take a discontinuous ring automorphism $alpha$ of $mathbb{C}$. For each $mathbb{C}$-vector space $V$, fix a basis and let $overline{alpha}_V : V to V$ be the function $(x_1,dots,x_n) mapsto (alpha(x_1),dots,alpha(x_n))$. Now we can construct an endofunctor $F$ of $textsf{FinVec}_mathbb{C}$ by sending each object to itself and a linear map $f : V to W$ to $overline{alpha}_W circ f circ overline{alpha}_V^{-1}$. This functor is not $C^0$ because (e.g.) the map on $text{Hom}(mathbb{C},mathbb{C})$ is canonically diffeo. to $alpha$.
– diracdeltafunk
Nov 28 '18 at 5:40
@QiaochuYuan To check my understanding: take a discontinuous ring automorphism $alpha$ of $mathbb{C}$. For each $mathbb{C}$-vector space $V$, fix a basis and let $overline{alpha}_V : V to V$ be the function $(x_1,dots,x_n) mapsto (alpha(x_1),dots,alpha(x_n))$. Now we can construct an endofunctor $F$ of $textsf{FinVec}_mathbb{C}$ by sending each object to itself and a linear map $f : V to W$ to $overline{alpha}_W circ f circ overline{alpha}_V^{-1}$. This functor is not $C^0$ because (e.g.) the map on $text{Hom}(mathbb{C},mathbb{C})$ is canonically diffeo. to $alpha$.
– diracdeltafunk
Nov 28 '18 at 5:40
1
1
@diracdeltafunk: it's unnecessary to fix a basis, and not quite. A vector space over $mathbb{C}$ is an abelian group $A$ together with an action $rho : mathbb{C} to text{End}(A)$. If $alpha : mathbb{C} to mathbb{C}$ is an endomorphism, then given an action $rho$ you can produce a new action $rho circ alpha$; this is the functor. Note that this construction doesn't require $alpha$ to be invertible.
– Qiaochu Yuan
Nov 28 '18 at 5:43
@diracdeltafunk: it's unnecessary to fix a basis, and not quite. A vector space over $mathbb{C}$ is an abelian group $A$ together with an action $rho : mathbb{C} to text{End}(A)$. If $alpha : mathbb{C} to mathbb{C}$ is an endomorphism, then given an action $rho$ you can produce a new action $rho circ alpha$; this is the functor. Note that this construction doesn't require $alpha$ to be invertible.
– Qiaochu Yuan
Nov 28 '18 at 5:43
1
1
@QiaochuYuan I'm familiar with the idea of constructing a functor $f^* : R-textsf{Mod} to S-textsf{Mod}$ from a ring homomorphism $f : S to R$. However, doesn't this functor send a linear map to itself (the same function of underlying sets)? If so, won't it give the identity function $text{Hom}(V,W) to text{Hom}(V,W)$ (which is necessarily continuous)?
– diracdeltafunk
Nov 28 '18 at 5:50
@QiaochuYuan I'm familiar with the idea of constructing a functor $f^* : R-textsf{Mod} to S-textsf{Mod}$ from a ring homomorphism $f : S to R$. However, doesn't this functor send a linear map to itself (the same function of underlying sets)? If so, won't it give the identity function $text{Hom}(V,W) to text{Hom}(V,W)$ (which is necessarily continuous)?
– diracdeltafunk
Nov 28 '18 at 5:50
3
3
Ooh, here's an endofunctor for the real case: compose any one of @QiaochuYuan 's wild endofunctors with the complexification functor (and then promptly forget the complex structure, of course)
– Monstrous Moonshiner
Nov 28 '18 at 6:44
Ooh, here's an endofunctor for the real case: compose any one of @QiaochuYuan 's wild endofunctors with the complexification functor (and then promptly forget the complex structure, of course)
– Monstrous Moonshiner
Nov 28 '18 at 6:44
|
show 4 more comments
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2
To clarify, over $mathbb{C}$ there are "wild" endofunctors coming from composing scalar multiplication with discontinuous endomorphisms $mathbb{C} to mathbb{C}$, of which there are many (because $mathbb{C}$ is algebraically closed). But $mathbb{R}$ has no nontrivial endomorphisms as a ring. All the endofunctors I know over $mathbb{R}$ are direct sums of Schur functors, which are $C^{infty}$ and even algebraic: en.wikipedia.org/wiki/Schur_functor
– Qiaochu Yuan
Nov 28 '18 at 4:17
1
@QiaochuYuan To check my understanding: take a discontinuous ring automorphism $alpha$ of $mathbb{C}$. For each $mathbb{C}$-vector space $V$, fix a basis and let $overline{alpha}_V : V to V$ be the function $(x_1,dots,x_n) mapsto (alpha(x_1),dots,alpha(x_n))$. Now we can construct an endofunctor $F$ of $textsf{FinVec}_mathbb{C}$ by sending each object to itself and a linear map $f : V to W$ to $overline{alpha}_W circ f circ overline{alpha}_V^{-1}$. This functor is not $C^0$ because (e.g.) the map on $text{Hom}(mathbb{C},mathbb{C})$ is canonically diffeo. to $alpha$.
– diracdeltafunk
Nov 28 '18 at 5:40
1
@diracdeltafunk: it's unnecessary to fix a basis, and not quite. A vector space over $mathbb{C}$ is an abelian group $A$ together with an action $rho : mathbb{C} to text{End}(A)$. If $alpha : mathbb{C} to mathbb{C}$ is an endomorphism, then given an action $rho$ you can produce a new action $rho circ alpha$; this is the functor. Note that this construction doesn't require $alpha$ to be invertible.
– Qiaochu Yuan
Nov 28 '18 at 5:43
1
@QiaochuYuan I'm familiar with the idea of constructing a functor $f^* : R-textsf{Mod} to S-textsf{Mod}$ from a ring homomorphism $f : S to R$. However, doesn't this functor send a linear map to itself (the same function of underlying sets)? If so, won't it give the identity function $text{Hom}(V,W) to text{Hom}(V,W)$ (which is necessarily continuous)?
– diracdeltafunk
Nov 28 '18 at 5:50
3
Ooh, here's an endofunctor for the real case: compose any one of @QiaochuYuan 's wild endofunctors with the complexification functor (and then promptly forget the complex structure, of course)
– Monstrous Moonshiner
Nov 28 '18 at 6:44