2 by 2 by 2 Tensors

We now study the case of 2\times2\times2 tensors over both the real and complex numbers. Our purpose is to give visualizations of theoretical work of others.

Visualizations over the Real numbers

We follow the work of Vin De Silva and Lek-Heng Lim in [DES-LIM2008], although several of these results appeared earlier in [KRUSKA1989]. They show that all 2\times2\times2 tensors are in S_3 but both S_2 and S_3\setminus S_2 have positive volume. The tensors in S_3\setminus S_2 can be divided into the cases D_3 and G_3.

Degenerate Separation Rank 3

The class D_3 consists of tensors not in S_2 that are a limit of tensors in S_2. The existence of tensors in D_3 shows that S_2 is not a closed set. A representative element of D_3 is

D_3 = [1,0]\otimes[1,0]\otimes[1,0]
+ [1,0] \otimes [0,1]\otimes [0,1]
+ [0,1]\otimes [1,0]\otimes [0,1]
\,,

which can also be viewed as

D_3 = \left[\begin{array}{cc}1&0\\0&0\end{array}\right]\otimes[1,0]
+ \left[\begin{array}{cc}0&1\\1&0\end{array}\right]\otimes [0,1]
\,.

Using the plane defined by the three tensors

T_1&= [1,0]\otimes [1,0]\otimes [1,0]\,, \\
T_2&=[1,0] \otimes [0,1]\otimes [0,1]\,,
\quad\text{and}\\
T_3&=[0,1]\otimes [1,0]\otimes [0,1]\,,

the point D_3 is in their center. Since these three tensors are in S_1, any tensor on the line connecting any two of them is in S_2. These lines help us identify the three tensors in the visualization, and thus help approximately locate the D_3 tensor.

Plotting S_2 in this way, we obtain the visualization

_images/D3.jpg

[data file 2x2x2pr2D3.dat] Notice that we obtain black points near D_3, and this lack of a void validates the result that there are tensors in S_2 near D_3. It does not allow us to distinguish points in D_3 from points in S_2.

General Separation Rank 3

The class G_3 consists of tensors not in S_2 that are not a limit of tensors in S_2. A representative element of G_3 is

G_3 = [1,1]\otimes[0,1]\otimes[0,1]
+ [1,-1]\otimes[1,0]\otimes[1,0]
+ [0,1]\otimes [1,1]\otimes [1,-1]
\,,

which can also be viewed as

G_3 = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]\otimes[1,0]
+ \left[\begin{array}{cc}0&1\\-1&0\end{array}\right]\otimes [0,1]
\,.

Using the plane defined by the three tensors

T_1&=[1,1]\otimes[0,1]\otimes[0,1]
\,,\\
T_2&=[1,-1]\otimes[1,0]\otimes[1,0]
\,,\quad\text{and}\\
T_3&=[0,1]\otimes [1,1]\otimes [1,-1]
\,,

the point G_3 is in their center. Since these three tensors are in S_1, any tensor on the line connecting any two of them is in S_2. These lines help us identify the three tensors in the visualization, and thus help approximately locate the G_3 tensor.

Plotting S_2 in this way, we obtain the visualization

_images/G3.jpg

[data file 2x2x2pr2G3.dat] We obtain black points everywhere outside of the triangle, and part of the region inside of the triangle. However, near G_3 there is a distinct void, indicating that this tensor cannot be approached as a limit of tensors in S_2.

We also note that we used only one-third the points to make this visualization as compared to the D_3 visualization, and yet obtained more black points. This indicates that the ALS routine had more difficulty on the tensors in the D_3 plane, perhaps because they are only limit points of S_2.

Random Example

Randomly choosing T_1, T_2, and T_3, we obtained a visualization

_images/2x2x2pr2random.jpg

[data file 2x2x2pr2random.dat] In this visualization we see both black regions and voids, suggesting that it is relatively easy to hit both types of behaviors.

Visualizations over the Complex numbers

Over the complex numbers all 2\times2\times2 tensors are in S_2. We illustrate this by considering the G_3 tensor defined above. Using the complex line defined by the two tensors

T_1&=[1,1]\otimes[0,1]\otimes[0,1]+[1,-1]\otimes[1,0]\otimes[1,0]
\,,\quad\text{and}\\
T_2&=[0,1]\otimes [1,1]\otimes [1,-1]
\,,

the point G_3 is the (real) midpoint of the line. Plotting S_2 in this way on [-1,2]\times[-i,i], we obtain the visualization

_images/G3C.jpg

[data file 2x2x2pr2G3C.dat] We obtain black points everywhere, indicating that G_3 is in S_2 over the complex numbers (prehaps as a limit point). The ALS routine had more trouble finding black points in the left half of the visualization, which is closer to T_2. This is counterintuitive, since T_2 is only rank 1 whereas T_1 is rank 2.