MHB A theorem on Quadratic Forms in Reid's Book not at all clear.

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The discussion revolves around a theorem on quadratic forms presented in Reid's book on Algebraic Geometry, which lacks a clear definition of quadratic forms. Participants clarify that a quadratic form can be defined as a function from a vector space to a field, determined by a bilinear form. There is confusion regarding Reid's notation, specifically the $x_i$'s in his theorem, which are identified as the coordinates of a vector in a specified basis. The conversation highlights that every quadratic form on a finite-dimensional vector space is diagonalizable and can be expressed in terms of symmetric bilinear forms. Overall, the thread emphasizes the need for clearer explanations in mathematical texts.
caffeinemachine
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Hello MHB,
I have been reading a book on Algebraic Geometry by Reid.

On page 15, there's a theorem on Quadratic forms. The book doesn't explicitly define what a Quadratic Form is. From Hoffman & Kunze's book on Linear Algebra I found that given an inner product space $V$ over a field $F$, the Quadratic Form determined by the inner product is a function from $V$ to $F$ which maps every vector $v\in V$ to the scalar $||v||^2$.

In the above context I can make sense of Theorem (B) in this:
https://docs.google.com/file/d/0B77QF0wgZJZ7VmhBcy1BaExTN28/edit

Does anybody see what Reid means by his Theorem B?
 
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caffeinemachine said:
From Hoffman & Kunze's book on Linear Algebra I found that given an inner product space $V$ over a field $F$, the Quadratic Form determined by the inner product is a function from $V$ to $F$ which maps every vector $v\in V$ to the scalar $||v||^2$.

That is a particular case of quadratic form. More general, $Q:V\to F$ is a quadratic form iff there exists a bilinear form $f:V\times V\to F$ such that $Q(x)=f(x,x)$ for all $x\in V.$ Note that a real inner product is a bilinear form.

Does anybody see what Reid means by his Theorem B?

Every cuadratic form on a finite dimensional vector space is diagonalizable. This, as a consequence that every quadratic form can be expressed as $Q(x)=f_s(x,x)$ with $f_s$ symmetric bilinear form (if $\operatorname{carac}F\neq 2$).
 
Thank you Fernando Revilla for your help. I have a few follow up questions.

Fernando Revilla said:
That is a particular case of quadratic form. More general, $Q:V\to F$ is a quadratic form iff there exists a bilinear form $f:V\times V\to F$ such that $Q(x)=f(x,x)$ for all $x\in V.$ Note that a real inner product is a bilinear form.
Is this the most general definition of a Quadratic Form?

Fernando Revilla said:
Every cuadratic form on a finite dimensional vector space is diagonalizable. This, as a consequence that every quadratic form can be expressed as $Q(x)=f_s(x,x)$ with $f_s$ symmetric bilinear form (if $\operatorname{carac}F\neq 2$).
To understand this better, can you please tell me what does Reid mean by $x_i$'s in his statement. I think his statement is incomplete. He says that there exists a basis such that $Q=\sum_{i=1}^n\varepsilon_ix_i^2$. He doesn't specify what are $x_i$'s.

Can you please write Reid's statement in a more intelligible form?

Thanks.
 
caffeinemachine said:
Is this the most general definition of a Quadratic Form?

Yes, that is the general definition.

To understand this better, can you please tell me what does Reid mean by $x_i$'s in his statement. I think his statement is incomplete. He says that there exists a basis such that $Q=\sum_{i=1}^n\varepsilon_ix_i^2$. He doesn't specify what are $x_i$'s.

Those $x_i$ are the coordinates of a generic vector $x\in V$ with respect to the mentioned basis, and $\epsilon_i$ are elementes of $F.$
 
caffeinemachine,

Having sat an examined reading course set by Professor Reid (Dr then) 20 years ago, based on this book, I can only wish you the best of luck in your quest. Thinking back to my revision from this book I am reminded of the following words from Dante's Inferno:

Abandon hope, all ye who enter here...(Sweating)
 
Semillon said:
caffeinemachine,

Having sat an examined reading course set by Professor Reid (Dr then) 20 years ago, based on this book, I can only wish you the best of luck in your quest. Thinking back to my revision from this book I am reminded of the following words from Dante's Inferno:

Abandon hope, all ye who enter here...(Sweating)
Haha! I know that feel bro!
 
Hi caffeinemachine,

It seems we have a common interest in learning differential geometry rigorously. I'm not sure if you were aware of this, but the following seems like a great resource for complementary free material that could provide alternative avenues when you get stuck. I'm currently using it to get my head around the various equivalent definitions of a manifold (e.g. Munkres vs Spivak).

Free Differential Geometry Books Download | Ebooks Online

Semillon.
 

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