Post History
#3: Post edited
by
peterh
·
2025-06-10T12:52:53Z (6 days ago)
- For the 2-index tensors, we have the unit matrix, or eye matrix ($M_{ab}=\delta_{ab}$).
- What is the case for more index tensors?
- Group operation would be that we consider the tensors as linear operations, and the multipliciation would be the concatenated execution of the operation.
- On a formalized way, for example for 3-index tensors, we are searching the 3-index tensor $M_{abc} \in \mathbb{R}^{3 \times 3 \times 3}$, for which
$$\forall_{M_{a'b'c'} \in \mathrm{R}^{3 \times 3 \times 3}}: M_{abc} \cdot M_{a'b'c'} = M_{a'b'c'}$$
- For the 2-index tensors, we have the unit matrix, or eye matrix ($M_{ab}=\delta_{ab}$).
- What is the case for more index tensors?
- Group operation would be that we consider the tensors as linear operations, and the multipliciation would be the concatenated execution of the operation.
- On a formalized way, for example for 3-index tensors, we are searching the 3-index tensor $M_{abc} \in \mathbb{R}^{3 \times 3 \times 3}$, for which
- $$\forall {M_{a'b'c'} \in \mathbb{R}^{3 \times 3 \times 3}}: M_{abc} \cdot M_{a'b'c'} = M_{a'b'c'}$$
#2: Post edited
by
peterh
·
2025-06-10T12:47:46Z (6 days ago)
- For the 2-index tensors, we have the unit matrix, or eye matrix ($M_{ab}=\delta_{ab}$).
- What is the case for more index tensors?
- Group operation would be that we consider the tensors as linear operations, and the multipliciation would be the concatenated execution of the operation.
On a formalized way, for example for 3-index tensors, we searching the 3-index tensor $M_{{abc} \in \mathbb{R}^{3 \times 3 \times 3}}$, for which- $$\forall_{M_{a'b'c'} \in \mathrm{R}^{3 \times 3 \times 3}}: M_{abc} \cdot M_{a'b'c'} = M_{a'b'c'}$$
- For the 2-index tensors, we have the unit matrix, or eye matrix ($M_{ab}=\delta_{ab}$).
- What is the case for more index tensors?
- Group operation would be that we consider the tensors as linear operations, and the multipliciation would be the concatenated execution of the operation.
- On a formalized way, for example for 3-index tensors, we are searching the 3-index tensor $M_{abc} \in \mathbb{R}^{3 \times 3 \times 3}$, for which
- $$\forall_{M_{a'b'c'} \in \mathrm{R}^{3 \times 3 \times 3}}: M_{abc} \cdot M_{a'b'c'} = M_{a'b'c'}$$
#1: Initial revision
by
peterh
·
2025-06-09T17:18:02Z (7 days ago)
What is the unit element in the space of more-than-2 indexed tensors?
For the 2-index tensors, we have the unit matrix, or eye matrix ($M_{ab}=\delta_{ab}$).
What is the case for more index tensors?
Group operation would be that we consider the tensors as linear operations, and the multipliciation would be the concatenated execution of the operation.
On a formalized way, for example for 3-index tensors, we searching the 3-index tensor $M_{{abc} \in \mathbb{R}^{3 \times 3 \times 3}}$, for which
$$\forall_{M_{a'b'c'} \in \mathrm{R}^{3 \times 3 \times 3}}: M_{abc} \cdot M_{a'b'c'} = M_{a'b'c'}$$