blow up is manifold?

[manic lintel]

I have no idea how to prove the \tilde{U_Y} is smooth complex submanifold. What do we use to show it?

[wispy tartan]

Heyyyy

We’re trying to show that U~Y⊂P^(k−1)×U is a smooth complex submanifold, right?

Do you remember what we usually check when we want to prove that a subset is a smooth complex submanifold?

I mean like what are the typical two conditions?

[manic lintel]

First, U~Y is zero set of a holomorphic map.
Second, Jacobian of the map is full rank. Is this right?

[wispy tartan]

That's exactly the 2 things we need

Now look at what the actual defining equations for U~Y are

Zjfi (x)−Zifj (x)=0 for all i,j≤k

Are these equations Holomorphic in both variables Z and x?

[manic lintel]

These equations are holomorphic on each local chart of P^k-1. So they are holomophic.

"On each local chart" means that for V_i := {z \in P^k-1 : z_i ≠ 0} , these equations are holomorphic on V_i × U.

[wispy tartan]

Exactlyyy

Since P^(k-1) is covered by open charts Vi and the equations are holomorphic on ecah Vi×U

The 2nd condition
Jacobian of the defining equations has full rank

At each point (Z,x)∈U~Y, what do the equations Zjfi(x) - zifj(x) =0
aactually constrain?

are they constraining Z or x or both?

what variable are they really restricting?

[manic lintel]

Actually, the equations constraning Z_i and x on V_i × U.

[wispy tartan]

If we fix a point x∈Y⊂U, then the values f1(x),…,fk(x) are just complex numbers

So for fixed x, this cuts out a projective line in P^(k−1), i.e. it’s constraining the Z's.

so actually, the equations mainly constrain Z, not x.
The x-dependence is parametric, meaning the equations vary holomorphically with x, but they don’t restrict x directly.

So now What do we need to check in the Jacobian to make sure that U~Y is smooth?

[manic lintel]

Ahh!! we think fixed (Z, x) \in U~Y!! So the equations are constraining the Z's , right?

So, we have to check that the Jacobian is rank k ?

[wispy tartan]

U got it dude

see since the equations are linear in Z and holomorphic in x, we can look at the Jacobian matrix of these equations with respect to all the variables — both the homogeneous coordinates Z (in affine chart) and the local coordinates of x∈U.

Are the defining equations independent at each point?

If the rank of jacobian equals the number of equations, then the level set is smooth and has the correct dimension.

[manic lintel]

Sorry, I can't judge the independence of equations....

At each point means at each (Z, x) \in U~Y ?

So, the Jacobian is rank 2k ?

[wispy tartan]

Yeah! At each point means each(Z,x)∈U~Y.

the Jacobian isn't rank 2𝑘

[manic lintel]

Ahh... Sorry, how to compute the rank of Jacobian?

Simply we calculate
∂(Z_if_j - Z_jf_i)/∂x_l and ∂(Z_if_j - Z_jf_i)/∂Z_k for all k, l ?

[wispy tartan]

Exactly u r thinking in right way

[manic lintel]

So, the Jacobian is k(k-1)/2 × (k+n) matrix ?

No, the Jacobian is k^2 × (k+n) matrix？

[wispy tartan]

Ah I see what you're thinking

Ur actually really close

1st we are looking at the equation Zjfi(x) - Zifj(x)=0 for all  1≤i<j≤k

[manic lintel]

Oh! So the Jacobian is k(k-1)/2 × (k+n) matrix?

[wispy tartan]

Yessss

[manic lintel]

But I have no idea how to compute the rank of Jacobian.

[wispy tartan]

Instead of jumping straight to computing the full Jacobian, let’s do a small concrete example to see how the structure works. Can we?

[manic lintel]

ok!

[wispy tartan]

Let's take k=2 and just one f1(x), f2(x) then defining equation is

Z2f1(x) - z1f2(x)=0

This is just one equation
so the Jacobian will be a 1 × (2 + n) matrix (2 for Z1,Z2, and n for the x-variables).

Now compute

∂/∂Z1 =−f2(x)
∂/∂Z2=f1(x)

∂/∂xl=Z2(∂f1/∂xl) −Z1(∂f2/∂xl)

[manic lintel]

The rank is almost 1? But, are there points in U~Y such that all entires of Jacobian are 0 ?

[wispy tartan]

So yeah
The Jacobian could have all entries zero only if
all fi(x) = 0 ---> so x∈Y
also dfi(x)=0 --> which would mean the defining functions are not transverse just like badly behaved

but in our 1st setup, the fi define Y as a smooth submanifold, so their differentials dfi are linearly independent at points in Y

so even if the Z-derivatives vanish (bcz of all fi(x)=0), the x-derivatives won't vanish, they're still active

So the jacobian never drops to rank 0. It always has full rank = number of equations and That's why U~Y is smooth

[manic lintel]

I apologize for the basic question, What is the relation linearly independence of df_1, ..., df_k and ∂/∂Z_i ?

[wispy tartan]

No need to apologize at all

So here's the idea:

The ∂/∂𝑍𝑖 derivatives only touch the Z’s, so they treat the
𝑓𝑖(𝑥) like constants
→ These vanish only when all
𝑓𝑖(𝑥)=0 (i.e.𝑥∈𝑌)

On the other hand, the
𝑑𝑓𝑖's are about how the 𝑓𝑖 change with respect to 𝑥
→ They appear in the
∂/∂𝑥ℓ parts of the Jacobian

→ These can still be nonzero even when 𝑓𝑖(𝑥)=0

So even if the Z-part of the jacobian vanishes (bcx you're on Y), the x-part, coming from dfi. can still be active

that's what prevents the full jacobian from dropping to zero rank

[manic lintel]

Ahh!!! I understand!!! If the Jacobian is rank 0 matrix, then all ∂/Z_i = f_j and ∂/∂x_l = ∂f_j/∂x_l are 0 . so all df_1, ..., df_k are zero. this is contradiction df_i are linearly independent, right?

[wispy tartan]

Yess Exactly

[manic lintel]

Since the rank of Jacobians is k(k-1)/2 , (And this is full rank. ) U~Y is smooth submanifold of P^k-1 × U!!!!

[wispy tartan]

YEAHHHH U DID IT!!!!!!

Now one last step
Can u figure out what the complex dimension of U~Y is?

[manic lintel]

k(k-1)/2 - (k-1 +n)?

no that is false, but I can't calculate correctly that...

[wispy tartan]

U r not completely wrong

Here the answer is
(k-1)+n -k(k-1)/2

[manic lintel]

Ahh!!!! I'm confused domain and codomain!!

[wispy tartan]

Let me explain it too

We’re looking at the map:

So

Domain= P^(k-1)xU
in local coordinates (Z1,......,Zk-1;x1,......,xn), so dimension = (k-1)+n

Codomain= C^(k(k-1)/2), since there are that many independent equations

Jacobian is a matrix of shape

[manic lintel]

I exactly understand!!

[shadow finch]

pre-uni maths and helping out with differential geometry.. that's insane