About invariant of quadratic space

Question

The following excerpt is from Scharlau: Quadratic and Hermitian Forms, pages 35-36.

One associates "in-variants" with the quadratic space so that the space is determined by its invariants as completely as possible. What is meaning of this statement? Also I know that dimension is invariant property for vector space and it is equal or greater than 0 so we can construct Grothendieck ring of integers $Z$. But why hyperbolic space has Grothendieck ring $Z/2Z$? How to construct it from hyperbolic space? Also someone explain ideal part which is totally confusing for me?

Answer 1

An invariant is a property that is preserved under certain transformations. In particular if $\mathcal{O}$ is a class of objects (more formally a category), $\mathcal{X}$ is a class of possible values and $P:\mathcal{O}\to\mathcal{X}$ is a property then we say that $P$ is an invariant if $P(X)=P(Y)$ whenever $X$ is isomorphic to $Y$.

Examples of invariants

• In the class of all groups: $P$ given by $P(G)=|G|$ (i.e. the cardinality of $G$) is an invariant. For short cardinality is an invariant for groups.
• In the class of all topological spaces: $P$ given by $P(X)=\begin{cases}\text{true} &\text{if }X\text{ is compact} \\ \text{false} &\text{otherwise}\end{cases}$. For short compactness is an invariant for topological spaces
• In the class of all vector spaces over a field $k$: $P$ given by $P(V)=\dim_k(V)$ is an invariant.

The last example is especially important because it is actually a complete invariant (the terminology is not standard), meaning the converse is also true: if $\dim_k(V)=\dim_k(W)$ then $V$ is isomorphic to $W$. Note that other two examples are not complete.

Counterexample: In the class of all metric spaces the property of $M$ being a complete metric space is not an invariant. Indeed, $\mathbb{R}$ is homeomorphic to $(0,1)$ (both with Euclidean metric) but the first one is a complete metric space while the other one is not.

Invariants are important because generally they are easier to calculate then raw classes of isomorphisms. They are tools used for a classification of objects. The goal is to obtain a list of invariants that fully describe an object just like the dimension fully describes a vector space.

Also note the subtlety: the dimension is an invariant for vector spaces equiped with a symmetric bilinear form (recall the definition of the Witt ring) but it is not a complete invariant here. Even though it is a complete invariant for raw vector spaces.

But why hyperbolic space has Grothendieck ring $Z/2Z$?

That's not what it says. That doesn't even make sense. For a Grothendieck ring you need a semiring to begin with. Anyway what it says is that there's a ring homomorphism

$$e:W(K)\to\mathbb{Z}_2$$ $$e([V])=\dim(V)\text{ mod }2$$

Formally I should write $$e([[V], [W]])=\dim(V)-\dim(W)\text{ mod }2$$ because there are two equivalence relations: one for the Witt ring and one for the associated Grothendieck ring. I'm writing $[V]=[[V],[0]]$ to simplify things.

This is well defined because if $V\sim W$ then by the definition $W$ can be obtained from $V$ by adding hyperbolic spaces. And the dimension of a hyperbolic space is always even meaning that $V$ and $W$ both have either even or odd dimension (even though they can be different). And thus taking ($\text{mod }2$) yields the same value for both. And it's a ring homomorphism because $\dim$ is a semiring homomorphism.