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<div class="slide" group="theme1">
<h1>Scalar Product and Metric Tensor</h1>
<div>What are the "basis
vectors" of whom the coordinates are the coefficients?</div>
<ol>
<li>To answer this, we need one more conceptual ingredient.</li>
<li>Linear functionals: A linear function from a vector space to its
field.
$$a = \mathbf{f}(\vec{x}) x \in V a \in \R$$
linearity:
$$\mathbf{f}(a \vec{x}) = a \mathbf{f}(\vec{x})$$
$$\mathbf{f}(\vec{x} + \vec{y}) = \mathbf{f}(\vec{x}) + \mathbf{f}(\vec{y})$$
</li>
<li>From the coordinate representation of $\vec{x}$ this leads
naturally to a coordinate representation of $\mathbf{f}$, like the
one we
saw for $g$:
$$\mathbf{f}(\vec{x}) = \mathbf{f}(x^{i} \vec{e}_{i})$$
$$\mathbf{f}(\vec{x}) = x^{i} \mathbf{f}(\vec{e}_{i})$$
$$\mathbf{f}(\vec{x}) = x^{i} f_{i}$$
where
$$f_i := \mathbf{f}(\vec{e}_{i})$$
(Einstein notation again.) We define $f_{i}$ to be teh coordinates
of the functional. The question is, again, in what basis?
</li>
</ol>
</div>
<div class="slide" group="theme1">
<h1>Scalar Product and Metric Tensor</h1>
<ol>
<li>There are many different such functions of course. Now, we define
combination/construction rules for any of them:
$$(a\mathbf{f})(\vec{x}) := a \mathbf{f}(\vec{x})$$
$$(\mathbf{f} + \mathbf{h})(\vec{x}) := \mathbf{f}(\vec{x}) + \mathbf{h}(\vec{x})$$</li>
<li>
With these rules, the linear functionals of a vector space become a
vector space themselves. It's called the <b>Co Vector Space</b> of
the underlying vector space. Both share the same scalar field.</li>
<li>Like for any vector space, the observations about linear
independence apply, and lead to the existence of coordinates and
bases.</li>
</ol>
</div>
<div class="slide" group="theme1">
<h1>Scalar Product and Metric Tensor</h1>
<ol>
<li>We have seen another way to construct coordinates for
covectors. How do we align the two? What is the basis that
corresponds to the coordinates we defined above? Let's try:
$$\mathbf{f} := f_{i} \mathbf{f}^{i}$$
<li>we apply this to a vector:
$$\mathbf{f}(\vec{v}) = (f_{i} \mathbf{f}^{i})(x^j \vec{x}_{j})$$
<li>from linearity of the $\mathbf{f}^{i}$ follows:
$$\mathbf{f}(\vec{v}) = f_{i} x^j \mathbf{f}^{i}(\vec{x}_{j})$$
<li>Now, let's pick the basis $\mathbf{f}^{i}$ such that
$$\mathbf{f}^{i}(\vec{x}_{j}) = \delta^i_j$$
<li>If we pick these bases, then the application of a functional to a
vector becomes the Einstein sum of the products of their
coordinates: $$\mathbf{f}(\vec{v}) = f_{i} x^j \delta^i_j = f_{i}
x^i$$
</li>
</ol>
</div>
<div class="slide" group="theme1">
<h1>Scalar Product and Metric Tensor</h1>
<div>Finally define the tensor product</div>
<ol>
<li>We have seen another way to construct coordinates for
covectors. How do we align the two? What is the basis that
corresponds to the coordinates we defined above? Let's try:
$$\mathbf{f} := f_{i} \mathbf{f}^{i}$$
<li>we apply this to a vector:
$$\mathbf{f}(\vec{v}) = (f_{i} \mathbf{f}^{i})(x^j \vec{x}_{j})$$
<li>from linearity of the $\mathbf{f}^{i}$ follows:
$$\mathbf{f}(\vec{v}) = f_{i} x^j \mathbf{f}^{i}(\vec{x}_{j})$$
<li>Now, let's pick the basis $\mathbf{f}^{i}$ such that
$$\mathbf{f}^{i}(\vec{x}_{j}) = \delta^i_j$$
<li>If we pick these bases, then the application of a functional to a
vector becomes the Einstein sum of the products of their
coordinates: $$\mathbf{f}(\vec{v}) = f_{i} x^j \delta^i_j = f_{i}
x^i$$
</li>
</ol>
</div>
<div class="slide" group="theme1">
<h1>Duality and Isomorphism</h1>
TANGENT
<ol>
<li>Like for any vector space, we can define linear functionals on the
co vector space too.
$$\xi(f) = a$$
How can such a functional be constructed? For example, pick one
vector $\vec{x}$ from the vector space, and define a functional:
$$\xi(f) := f(\vec{x})$$
It turns out <b>every</b> linear functional of the co vector space
can be written in this way. There is a 1:1 correspondence between
the linear functionals of the co vector space and the vectors of its
vector space.</li>
<li>Thus, we say that the co vector space of the co vector space
<b>is</b> the vector space again.</li>
<li>Btw. this introduces two important concepts: <b>Isomorphism</b>
and <b>Duality</b>.</li>
<li>Isomorphism is also how we got to think of vectors as <b>the
same</b> as their coordinate tuples: Every vector space of
dimension $n$ is — not the same as! but: — <b>isomorphic</b> to
$\R^{n}$ with regard to the vector space operations addition and
scalar multiplication.</li>
</ol>
</div>
<div class="slide noprint" group="theme2" name="theme2">
<div class="banner">
<div class="content">
II - The Dimension of Tensor Spaces (Dimensionality of the vector spaces of higher order tensors.)
</div>
</div>
</div>
<div class="slide print" group="theme2">
<h1>Code Snippets</h1>
<div class="snippet">
<div class="label">state_base.js</div>
<pre class="prettyprint lang-scm">
(friends '(of) '(parentheses))
function $(id) {
return document.getElementById(id);
}
</pre>
<pre class="prettyprint lang-js">
(friends '(of) '(parentheses))
function $(id) {
return document.getElementById(id);
}
</pre>
</div>
</div>
<div class="slide" group="theme2">
<h1>Game Plan</h1>
<p></p>
</div>
<div class="slide" group="theme2">
<h1>Conclusion</h1>
<p></p>
</div>
<div class="slide noprint" group="theme3" name="theme3">
<div class="banner">
<div class="content">
III - Tensors in Tensorflow
</div>
</div>
</div>
<div class="slide" group="theme3">
<h1>Conclusion</h1>
<p></p>
</div>