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I don't understand what does $\phi_I$ mean

The proof includes writing $\phi_I$ as a product of $\phi_{i_1}\phi_{i_2},\dots$, but it doesn't explain what the LHS really means

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    $\begingroup$ It's just the name they used for the tensor. They could have called it "Fred" if they had wanted to. $\endgroup$ – Carl Mummert Jun 2 '14 at 2:11
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    $\begingroup$ It's just a name. For each $k$-tuple of integers $I$ from the set $\{1, \dots, n\}$, there is a tensor which we choose to write $\phi_I$ to indicate that it corresponds to the $k$-tuple $I$. $\endgroup$ – Michael Albanese Jun 2 '14 at 2:11

$\phi_I$ is a $k$-multilinear form. This means $\phi_I$ takes in a $k$-tuple of vectors and is linear in slot if you fix the others. For example: $$\phi_I(c{\bf v}+{\bf w},{\bf v}_2,\dots,{\bf v}_k) = c\phi_I({\bf v},{\bf v}_2,\dots,{\bf v}_k)+ \phi_I({\bf w},{\bf v}_2,\dots,{\bf v}_k)$$ where $c$ is a scalar and the v's are vectors. Likewise, $\phi_I$ is linear in the other arguments. By "form" I mean that the output: $\phi_I({\bf v}_1,\dots,{\bf v}_n)$ is a scalar ($\phi_I$ maps $k$-tuples of vectors to scalars).

Now $I$ is a $k$-tuple of integers. Given a list of vectors from the given basis: $a_{j_1},\dots,a_{j_k}$ (repeats allowed), $\phi_I(a_{j_1},\dots,a_{j_k})$ spits out 0 unless $(j_1,\dots,j_k)=I$ (the subscripts match identically).

For example: $\mathbb{R}^2$ has the basis ${\bf i}=(1,0)$ and ${\bf j}=(0,1)$. Then... $$\phi_{(1,1)}({\bf i},{\bf i})=1, \quad \phi_{(1,1)}({\bf i},{\bf j})=0, \quad \phi_{(1,1)}({\bf j},{\bf i})=0, \quad \phi_{(1,1)}({\bf j},{\bf j})=0$$ $$\phi_{(1,2)}({\bf i},{\bf i})=0, \quad \phi_{(1,2)}({\bf i},{\bf j})=1, \quad \phi_{(1,2)}({\bf j},{\bf i})=0, \quad \phi_{(1,2)}({\bf j},{\bf j})=0$$ $$\phi_{(2,1)}({\bf i},{\bf i})=0, \quad \phi_{(2,1)}({\bf i},{\bf j})=0, \quad \phi_{(2,1)}({\bf j},{\bf i})=1, \quad \phi_{(2,1)}({\bf j},{\bf j})=0$$ $$\phi_{(2,2)}({\bf i},{\bf i})=0, \quad \phi_{(2,2)}({\bf i},{\bf j})=0, \quad \phi_{(2,2)}({\bf j},{\bf i})=0, \quad \phi_{(2,2)}({\bf j},{\bf j})=1$$

Also, for example, $$\phi_{(1,2)}((1,2),(3,4)) = \phi_{(1,2)}({\bf i}+2{\bf j},3{\bf i}+4{\bf j}) =\phi_{(1,2)}({\bf i},3{\bf i}+4{\bf j})+2\phi_{(1,2)}({\bf j},3{\bf i}+4{\bf j})$$ $$=3\phi_{(1,2)}({\bf i},{\bf i})+4\phi_{(1,2)}({\bf i},{\bf j})+6\phi_{(1,2)}({\bf j},{\bf i})+8\phi_{(1,2)}({\bf j},{\bf j})=3(0)+4(1)+6(0)+8(0)=4$$

  • $\begingroup$ Interesting, from your example with $V = \mathbb{R}^2$, I take it that this means $\phi_I$ must take on $\dim V$ vectors? So that $i_k = n$ always? $\endgroup$ – Hawk Jun 4 '14 at 4:12
  • $\begingroup$ $V=\mathbb{R}^2$, but no, it wasn't necessary to have a pair of vectors as inputs. Any number of vector inputs works as well. I chose 2 so the number of terms to deal with wasn't too rediculous. $\endgroup$ – Bill Cook Jun 5 '14 at 13:32
  • $\begingroup$ Just keep in mind that the number of input vectors must match the length of $ I$. $\endgroup$ – Bill Cook Jun 5 '14 at 13:34
  • $\begingroup$ So something like $\phi_{(2,1)}(\mathbf{i})$ is well-defined? $\endgroup$ – Hawk Jun 5 '14 at 23:21
  • $\begingroup$ No. $\varphi_{(2,1)}$ requires 2 inputs because $(2,1)$ has 2 coordinates. Something like $\varphi_{(2)}$ would only take 1 input such as: $\varphi_{(2)}(5{\bf i}-3{\bf j})=-3$ since $\varphi_{(2)}({\bf i})=0$ and $\varphi_{(2)}({\bf j})=1$. $\endgroup$ – Bill Cook Jun 6 '14 at 3:43

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