A linear form is a linear map from a vector space to a scalar from the vector space’s underlying field.

\(\hom (V, F)\)

Linear forms can be represented as matrix operators.

\(v^TM=f\)

Where \(M\) has only one column.

\(f(M)=f(v)\)

We introduce \(e_i\), the element vector. This is \(0\) for all entries except for \(i\) where it is \(1\). Any vector can be shown as a sum of these vectors multiplied by a scalar.

\(f(M)=f(\sum^m_{i=1}a_{i}e_i)\)

\(f(M)=\sum_{i=1}^mf(a_{i}e_i)\)

\(f(M)=\sum_{i=1}^ma_if(e_i)\)

\(f(M)=\sum_{i=1}^ma_if(e_i)\)

\(f(M)=\sum_{i=1}^ma_i\)

The dual space \(V^*\) of vector space \(V\) is the set of all linear forms, \(\hom(V,F)\).

\(v\in V\)

\(f\in F\)

\(av = f\)

\(bv = g\)

\((a\oplus b)v=f+g\)

\((a\oplus b)v=av + bv\)

So there is some operation we can do on two members of dual space

Linear in addition. That is, if we have two dual "things", we can define the addition of functions as the operation which results int he outputs being added.

what about linear in scalar? same approach.

Well we define

\((c\odot a)=cav\)

The dual space forms a vector space. We can define addition and scalar multiplication on members of the dual space.

The dimension of the dual space is the same as the underlying space.

We have defined the dual space. A vector in dual space will have also have components and a basis.

\(\mathbf w=\sum_i w_i f^j\)

So how we describe the components will depend on the choice of basis.

We choose the dual basis, the basis for \(V^*\) as:

\(\mathbf e_i \mathbf f^j =\delta_i^j\)

If the basis changes, so does the dual basis.

We write the dual basis as \(e^j\)

A bilinear form takes two vectors and produces a scalar from the underyling field.

This is in contrast to a linear form, which only has one input.

In addition, the function is linear in both arguments.

\(\phi (au+x, bv+y)=\phi (au,bv)+\phi (au,y)+\phi (x,bv)+\phi (x,y)\)

\(\phi (au+x, bv+y)=ab\phi (u,v)+a\phi (u,y)+b\phi (x,v)+\phi (x,y)\)

They can be represented as:

\(\phi (u,v)=v^TMu\)

\(f(M)=f([v_1,v_2])\)

We introduce \(e_i\), the element vector. This is \(0\) for all entries except for \(i\) where it is \(1\). Any vector can be shown as a sum of these vectors multiplied by a scalar.

\(f(M)=f([\sum^m_{i=1}a_{1i}e_i,\sum^m_{i=1}a_{2i}e_i])\)

\(f(M)=\sum_{k=1}^mf([a_{1k}e_k,\sum^m_{i=1}a_{2i}e_i])\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}f([a_{1k}e_k,a_{2i}e_i])\)

Because this in linear in scalars:

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}a_{2i}f([e_k,e_i])\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}a_{2i}e_k^TMe_i\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}a_{2i}e_k^Te_i\)

\(f(M)=\sum_{k=1}^m\sum^m_{i=1}a_{1k}a_{2i}\delta_i^k\)

\(f(M)=\sum^m_{i=1}a_{1i}a_{2i}\)

\(v^TMu=f\)

If the operator is \(I\) then we have the dot product.

\(v^Tu\)

Given a metric \(M\), two vectors \(v\) and \(u\) are orthogonal if:

\(v^TMu=0\)

For example if we have the metric \(M=I\), then two vectors are orthogonal if:

\(v^Tu=0\)

If we have a bilinear form we can write the form as:

\(u^TMv\)

After a transformation \(P\) to the vectors it is:

\((Pu)^TM(Pv)\)

\(u^TP^TMPv\)

So the value of the metric will be unaffected if:

\(u^TP^TMPv=u^TMv\)

\(P^TMP=M\)

Different metrics can produce the same group. For example multiplying the metric by a constant.

\(P^TMP=M\)

The bilinear form is:

\(u^TMv\)

The transformations which preserve this are:

\(P^TMP=M\)

If the metic is \(M=I\) then the condition is:

\(P^TP=I\)

\(P^T=P^{-1}\)

These form the orthogonal group.

We use \(O\) instead of \(P\):

\(O^T=O^{-1}\)

The orthogonal group is the rotations and reflections.

The orthogonal group depends on the dimension of the vector space, and the underlying field. So we can have:

\(O(n, R)\); and

\(O(n, C)\).

\(O(n)\) means \(O(n,R)\).

The generally refer to the reals only.

The special orthogonal group, \(SO(n,F)\), is the subgroup of the orthogonal group where \(|M|=1\).

As a result it includes only the rotation operators, not the flip operators.

\(SO(3)\) is rotations in 3d space.

\(SO(2)\) is rotations in 2d space.

The orthogonal group has determinants of \(-1\) or \(1\).

\(O^T=O^{-1}\)

\(\det (O^T)=\det (O^{-1})\)

\(\det O=\dfrac{1}{\det O}\)

\(\det O=\pm 1\)

The special linear group, \(SL(n,F)\), is the subgroup of \(GL(n,F)\) where the determinants are \(1\).

That is, \(|M|=1\)

These are endomorphisms, not forms.