Mixing
Tangent space to an affine variety $X subset mathbb{A}^n$ isomorphic to affine space $mathbb{A}^{n-1}$
$begingroup$
Let $mathbb{A}^n = mathbb{A}^n_{mathbb{C}}$ be n-affine over the field of complex numbers.
Let $f in mathbb{C}[x_1, cdots, x_n]$ be an irreducible, non-constant polynomial, and let $X = mathbb{V}(f)$ be the vanishing set of $f$.
For a point $p = (p_1, cdots, p_n)in X$, define the tangent space $$T_pX := {q = (q_1, cdots, q_n) in mathbb{A}^n | sum_{i=1}^nfrac{partial f}{partial x_i}|_p (q_i - p_i) = 0}.$$
A point $p in X$ is singular if $frac{partial f}{partial x_i}|_p = 0$ for all $1 leq i leq n$, and non-singular otherwise.
Why is it that if $pin X$ is non-singular, then $T_pX simeq mathbb{A}^{n-1}$?
algebraic-geometry
asked Jan 2 at 13:15
38917803891780
816
$endgroup$
add a comment |
$begingroup$
Let $mathbb{A}^n = mathbb{A}^n_{mathbb{C}}$ be n-affine over the field of complex numbers.
Let $f in mathbb{C}[x_1, cdots, x_n]$ be an irreducible, non-constant polynomial, and let $X = mathbb{V}(f)$ be the vanishing set of $f$.
For a point $p = (p_1, cdots, p_n)in X$, define the tangent space $$T_pX := {q = (q_1, cdots, q_n) in mathbb{A}^n | sum_{i=1}^nfrac{partial f}{partial x_i}|_p (q_i - p_i) = 0}.$$
A point $p in X$ is singular if $frac{partial f}{partial x_i}|_p = 0$ for all $1 leq i leq n$, and non-singular otherwise.
Why is it that if $pin X$ is non-singular, then $T_pX simeq mathbb{A}^{n-1}$?
algebraic-geometry
asked Jan 2 at 13:15
38917803891780
816
$endgroup$
add a comment |
$begingroup$
Let $mathbb{A}^n = mathbb{A}^n_{mathbb{C}}$ be n-affine over the field of complex numbers.
Let $f in mathbb{C}[x_1, cdots, x_n]$ be an irreducible, non-constant polynomial, and let $X = mathbb{V}(f)$ be the vanishing set of $f$.
For a point $p = (p_1, cdots, p_n)in X$, define the tangent space $$T_pX := {q = (q_1, cdots, q_n) in mathbb{A}^n | sum_{i=1}^nfrac{partial f}{partial x_i}|_p (q_i - p_i) = 0}.$$
A point $p in X$ is singular if $frac{partial f}{partial x_i}|_p = 0$ for all $1 leq i leq n$, and non-singular otherwise.
Why is it that if $pin X$ is non-singular, then $T_pX simeq mathbb{A}^{n-1}$?
algebraic-geometry
asked Jan 2 at 13:15
38917803891780
816
$endgroup$
Let $mathbb{A}^n = mathbb{A}^n_{mathbb{C}}$ be n-affine over the field of complex numbers.
Let $f in mathbb{C}[x_1, cdots, x_n]$ be an irreducible, non-constant polynomial, and let $X = mathbb{V}(f)$ be the vanishing set of $f$.
For a point $p = (p_1, cdots, p_n)in X$, define the tangent space $$T_pX := {q = (q_1, cdots, q_n) in mathbb{A}^n | sum_{i=1}^nfrac{partial f}{partial x_i}|_p (q_i - p_i) = 0}.$$
A point $p in X$ is singular if $frac{partial f}{partial x_i}|_p = 0$ for all $1 leq i leq n$, and non-singular otherwise.
Why is it that if $pin X$ is non-singular, then $T_pX simeq mathbb{A}^{n-1}$?
algebraic-geometry
algebraic-geometry
asked Jan 2 at 13:15
38917803891780
816
asked Jan 2 at 13:15
38917803891780
816
asked Jan 2 at 13:15
38917803891780
816
asked Jan 2 at 13:15
38917803891780
816
asked Jan 2 at 13:15
38917803891780
816
816
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I assume you are not using Schemes. If $p$ is non-singular then the matrix $(frac{partial f}{partial x_1} ~dots~ frac{partial f}{partial x_n})$ has full rank, so its kernel (which is exactly $T_p X - p$ is a $n-1$ dimensional subspace of $mathbb{C}^n = mathbb{A}^n$ (and therefore isomorphic to $mathbb{A}^{n-1}$).
answered Jan 2 at 13:22
0x5390x539
1,067317
$endgroup$
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
|
show 6 more comments
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1 Answer
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1 Answer
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$begingroup$
I assume you are not using Schemes. If $p$ is non-singular then the matrix $(frac{partial f}{partial x_1} ~dots~ frac{partial f}{partial x_n})$ has full rank, so its kernel (which is exactly $T_p X - p$ is a $n-1$ dimensional subspace of $mathbb{C}^n = mathbb{A}^n$ (and therefore isomorphic to $mathbb{A}^{n-1}$).
answered Jan 2 at 13:22
0x5390x539
1,067317
$endgroup$
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
|
show 6 more comments
$begingroup$
I assume you are not using Schemes. If $p$ is non-singular then the matrix $(frac{partial f}{partial x_1} ~dots~ frac{partial f}{partial x_n})$ has full rank, so its kernel (which is exactly $T_p X - p$ is a $n-1$ dimensional subspace of $mathbb{C}^n = mathbb{A}^n$ (and therefore isomorphic to $mathbb{A}^{n-1}$).
answered Jan 2 at 13:22
0x5390x539
1,067317
$endgroup$
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
|
show 6 more comments
$begingroup$
I assume you are not using Schemes. If $p$ is non-singular then the matrix $(frac{partial f}{partial x_1} ~dots~ frac{partial f}{partial x_n})$ has full rank, so its kernel (which is exactly $T_p X - p$ is a $n-1$ dimensional subspace of $mathbb{C}^n = mathbb{A}^n$ (and therefore isomorphic to $mathbb{A}^{n-1}$).
answered Jan 2 at 13:22
0x5390x539
1,067317
$endgroup$
I assume you are not using Schemes. If $p$ is non-singular then the matrix $(frac{partial f}{partial x_1} ~dots~ frac{partial f}{partial x_n})$ has full rank, so its kernel (which is exactly $T_p X - p$ is a $n-1$ dimensional subspace of $mathbb{C}^n = mathbb{A}^n$ (and therefore isomorphic to $mathbb{A}^{n-1}$).
answered Jan 2 at 13:22
0x5390x539
1,067317
edited Jan 2 at 15:01
answered Jan 2 at 13:22
0x5390x539
1,067317
answered Jan 2 at 13:22
0x5390x539
1,067317
answered Jan 2 at 13:22
0x5390x539
1,067317
1,067317
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
|
show 6 more comments
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
Thank you for your answer. My linear algebra needs some serious brushing up. As I remember, if $A in M_n(K)$ is a matrix, then the kernel of A $$text{ker}(A) = { v in K^n | Av = 0}.$$ If we fix a vector $pin K^n$, then is this set $${v in K^n | A(p-v) = 0}$$ also a kernel of some matrix?
$endgroup$
– 3891780
Jan 2 at 14:58
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
You're right. You have to move the kernel by $p$. I'll edit my answer.
$endgroup$
– 0x539
Jan 2 at 15:00
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
Wait, $p$ will not be in the kernel of $A$ in general (that wouldn't make any sene since $p$ is a point and not a vector).
$endgroup$
– 0x539
Jan 2 at 15:03
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
In our case $p$ is an element of the tangent space, so $$ T_pX = {q in mathbb{A}^n | (frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p(q- p) = 0 } = text{ker}(frac{partial f}{partial x_1} , cdots, frac{partial f}{partial x_n})|_p $$
$endgroup$
– 3891780
Jan 2 at 15:10
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
$begingroup$
@3891780 $q - p$ is in the kernel, but you only want $q$ for the tangent space.
$endgroup$
– 0x539
Jan 2 at 15:13
|
show 6 more comments
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