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Appendix B Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Basics of the Tensor Product \(S \otimes T\)

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Problem: Expand the tensor product \(S \otimes T\) for \(S = 3e_1 - e_2\) and \(T = e_1 + 4e_2\).

Applying the distributive law:

\[S \otimes T = (3e_1 - e_2) \otimes (e_1 + 4e_2)\]
\[= 3 \cdot 1\,e_1 \otimes e_1 + 3 \cdot 4\,e_1 \otimes e_2 + (-1) \cdot 1\,e_2 \otimes e_1 + (-1) \cdot 4\,e_2 \otimes e_2\]
\[\boxed{S \otimes T = 3\,e_1 \otimes e_1 + 12\,e_1 \otimes e_2 - e_2 \otimes e_1 - 4\,e_2 \otimes e_2}\]

Verification: The component matrix is \(\begin{pmatrix} 3 & 12 \\ -1 & -4 \end{pmatrix}\). This matches the outer product of the column vector \(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\) and the row vector \(\begin{pmatrix} 1 & 4 \end{pmatrix}\). ✓

B-2. Non-commutativity of Tensor Products

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Problem: Expand \(T \otimes S\) (where \(T = e_1 + 4e_2\), \(S = 3e_1 - e_2\)) and verify the difference from \(S \otimes T\).

\[T \otimes S = (e_1 + 4e_2) \otimes (3e_1 - e_2)\]
\[= 1 \cdot 3\,e_1 \otimes e_1 + 1 \cdot (-1)\,e_1 \otimes e_2 + 4 \cdot 3\,e_2 \otimes e_1 + 4 \cdot (-1)\,e_2 \otimes e_2\]
\[\boxed{T \otimes S = 3\,e_1 \otimes e_1 - e_1 \otimes e_2 + 12\,e_2 \otimes e_1 - 4\,e_2 \otimes e_2}\]

Comparison:

Basis Coefficients of \(S \otimes T\) Coefficients of \(T \otimes S\)
\(e_1 \otimes e_1\) \(3\) \(3\)
\(e_1 \otimes e_2\) \(12\) \(-1\)
\(e_2 \otimes e_1\) \(-1\) \(12\)
\(e_2 \otimes e_2\) \(-4\) \(-4\)

The coefficients of \(e_1 \otimes e_2\) and \(e_2 \otimes e_1\) are swapped, confirming that \(S \otimes T \neq T \otimes S\). The tensor product is non-commutative. ✓

B-3. Linear Combinations of Tensors

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Problem: Find \(P + Q\) and \(2P - Q\).

\(P = 2\,e_1 \otimes e_1 - 3\,e_1 \otimes e_2 + e_2 \otimes e_1 + 5\,e_2 \otimes e_2\)

\(Q = -e_1 \otimes e_1 + 4\,e_1 \otimes e_2 - e_2 \otimes e_1 + 2\,e_2 \otimes e_2\)

\(P + Q\): Add the coefficients of corresponding basis elements.

\[P + Q = (2-1)\,e_1 \otimes e_1 + (-3+4)\,e_1 \otimes e_2 + (1-1)\,e_2 \otimes e_1 + (5+2)\,e_2 \otimes e_2\]
\[\boxed{P + Q = e_1 \otimes e_1 + e_1 \otimes e_2 + 7\,e_2 \otimes e_2}\]

\(2P - Q\):

\[2P - Q = (4-(-1))\,e_1 \otimes e_1 + (-6-4)\,e_1 \otimes e_2 + (2-(-1))\,e_2 \otimes e_1 + (10-2)\,e_2 \otimes e_2\]
\[\boxed{2P - Q = 5\,e_1 \otimes e_1 - 10\,e_1 \otimes e_2 + 3\,e_2 \otimes e_1 + 8\,e_2 \otimes e_2}\]

Verification: Confirm that \((P+Q) + (2P-Q) = 3P\). \(3P = 6\,e_1\otimes e_1 - 9\,e_1\otimes e_2 + 3\,e_2\otimes e_1 + 15\,e_2\otimes e_2\). Left-hand side: \((1+5)\,e_1\otimes e_1 + (1-10)\,e_1\otimes e_2 + (0+3)\,e_2\otimes e_1 + (7+8)\,e_2\otimes e_2 = 6\,e_1\otimes e_1 - 9\,e_1\otimes e_2 + 3\,e_2\otimes e_1 + 15\,e_2\otimes e_2\). Consistent. ✓

B-4. Dimensions of Tensor Spaces

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Problem: When \(V\) is 4-dimensional, find the dimensions of \(T^2(V)\) and \(T^3(V)\).

Since the dimension of \(T^r(V)\) is \(n^r\) when \(V\) is \(n\)-dimensional:

\[\boxed{\dim T^2(V) = 4^2 = 16, \qquad \dim T^3(V) = 4^3 = 64}\]

Verification: The basis of \(T^2(V)\) consists of \(e_i \otimes e_j\) (\(i,j = 1,2,3,4\)), giving \(4 \times 4 = 16\) elements. The basis of \(T^3(V)\) consists of \(e_i \otimes e_j \otimes e_k\), giving \(4 \times 4 \times 4 = 64\) elements. ✓

B-5. Expansion of Einstein Summation

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Problem: Write out \(S = S^{ij}\,e_i \otimes e_j\) (where \(V\) is 3-dimensional) using \(\Sigma\) notation, and determine the number of terms.

\[\boxed{S = \sum_{i=1}^{3}\sum_{j=1}^{3} S^{ij}\,e_i \otimes e_j}\]

Expanding explicitly:

\[S = S^{11}\,e_1\otimes e_1 + S^{12}\,e_1\otimes e_2 + S^{13}\,e_1\otimes e_3 + S^{21}\,e_2\otimes e_1 + S^{22}\,e_2\otimes e_2 + S^{23}\,e_2\otimes e_3$$ $$+ S^{31}\,e_3\otimes e_1 + S^{32}\,e_3\otimes e_2 + S^{33}\,e_3\otimes e_3\]

The number of terms is \(3 \times 3 = \boxed{9}\).

B-6. Violations of the Einstein Convention

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Problem: Identify which of the following violate the conventions of Einstein summation notation.

(a) \(A^i B^j\,e_i \otimes e_j\) — \(i\) appears once in \(A^i\) (upper) and once in \(e_i\) (lower); \(j\) appears once in \(B^j\) (upper) and once in \(e_j\) (lower). No violation. ✓

(b) \(A^i B^i C^i\) — The index \(i\) appears 3 times in the upper position. Convention violation. The same index must not appear more than twice in a single term.

(c) \(S^{ij}\,T_{jk}\) — \(j\) appears once upper (in \(S^{ij}\)) and once lower (in \(T_{jk}\)), forming a contraction. \(i\) appears once upper (free index), and \(k\) appears once lower (free index). No violation. ✓

(d) \(A^i B_j\) — \(i\) appears once upper (free index), \(j\) appears once lower (free index). No contraction occurs. No violation. ✓

(e) \(S^{ij}\,e_i \otimes e_i\) — The index \(i\) appears once upper (in \(S^{ij}\)), once lower (in \(e_i\)), and once more lower (in \(e_i\)), for a total of 3 occurrences. Convention violation. Additionally, \(j\) appears only once in the upper position with no corresponding lower index, making it inconsistent as either a free index or a contraction.

\[\boxed{\text{Violations: (b) and (e)}}\]

B-7. Reconstructing a Tensor from Components

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Problem: Given \(S^{11}=2,\; S^{12}=-1,\; S^{21}=0,\; S^{22}=3\), write \(S\) explicitly.

\[S = S^{ij}\,e_i \otimes e_j = \sum_{i=1}^{2}\sum_{j=1}^{2} S^{ij}\,e_i \otimes e_j\]
\[\boxed{S = 2\,e_1 \otimes e_1 - e_1 \otimes e_2 + 0\,e_2 \otimes e_1 + 3\,e_2 \otimes e_2 = 2\,e_1 \otimes e_1 - e_1 \otimes e_2 + 3\,e_2 \otimes e_2}\]

B-8. Tensor Product of Rank-1 and Rank-2 Tensors

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Problem: Expand the tensor product \(A \otimes S \in T^3(V)\) for \(A = 2e_1 + e_2\) and \(S = e_1 \otimes e_2 - 3\,e_2 \otimes e_1\).

\[A \otimes S = (2e_1 + e_2) \otimes (e_1 \otimes e_2 - 3\,e_2 \otimes e_1)\]

Applying the distributive law:

\[= 2e_1 \otimes (e_1 \otimes e_2) + 2e_1 \otimes (-3\,e_2 \otimes e_1) + e_2 \otimes (e_1 \otimes e_2) + e_2 \otimes (-3\,e_2 \otimes e_1)\]
\[= 2\,e_1 \otimes e_1 \otimes e_2 - 6\,e_1 \otimes e_2 \otimes e_1 + e_2 \otimes e_1 \otimes e_2 - 3\,e_2 \otimes e_2 \otimes e_1\]
\[\boxed{A \otimes S = 2\,e_1 \otimes e_1 \otimes e_2 - 6\,e_1 \otimes e_2 \otimes e_1 + e_2 \otimes e_1 \otimes e_2 - 3\,e_2 \otimes e_2 \otimes e_1}\]

Verification: \(T^3(V)\) has dimension \(2^3 = 8\). The result above is a linear combination of 4 basis vectors, which is consistent as an element of \(T^3(V)\). Checking the components, we should have \((A \otimes S)^{ijk} = A^i S^{jk}\). With \(A^1 = 2, A^2 = 1\) and \(S^{12} = 1, S^{21} = -3\) (all others zero): \((A\otimes S)^{112} = 2 \cdot 1 = 2\), \((A\otimes S)^{121} = 2 \cdot (-3) = -6\), \((A\otimes S)^{212} = 1 \cdot 1 = 1\), \((A\otimes S)^{221} = 1 \cdot (-3) = -3\). All match. ✓

Medium

M-1. Condition for Decomposable Tensors

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Problem: Show that the necessary and sufficient condition for \(W = a\,e_1\otimes e_1 + b\,e_1\otimes e_2 + c\,e_2\otimes e_1 + d\,e_2\otimes e_2\) to be a decomposable tensor is \(ad - bc = 0\).

Solution Strategy

Assuming \(W = S \otimes T\), compare components to derive the necessary condition \(ad = bc\). Conversely, when \(ad = bc\), explicitly construct \(S\) and \(T\).

Necessary Condition (\(W\) is decomposable \(\Rightarrow\) \(ad - bc = 0\))

Let \(S = \alpha e_1 + \beta e_2\) and \(T = \gamma e_1 + \delta e_2\). Then:

\[S \otimes T = \alpha\gamma\,e_1\otimes e_1 + \alpha\delta\,e_1\otimes e_2 + \beta\gamma\,e_2\otimes e_1 + \beta\delta\,e_2\otimes e_2\]

If \(W = S \otimes T\), comparing coefficients gives:

\[a = \alpha\gamma, \quad b = \alpha\delta, \quad c = \beta\gamma, \quad d = \beta\delta\]

Therefore:

\[ad = (\alpha\gamma)(\beta\delta) = \alpha\beta\gamma\delta$$ $$bc = (\alpha\delta)(\beta\gamma) = \alpha\beta\gamma\delta\]

Hence \(ad = bc\), i.e., \(ad - bc = 0\).

Sufficient Condition (\(ad - bc = 0\) \(\Rightarrow\) \(W\) is decomposable)

Assume \(ad - bc = 0\). We proceed by cases.

Case 1: When \(a = b = c = d = 0\). Then \(W = 0 = 0 \otimes 0\), which is decomposable.

Case 2: When at least one of \(a, b, c, d\) is nonzero.

  • If \(a \neq 0\): From \(ad = bc\), we have \(d = bc/a\).

Set: $\(S = a\,e_1 + c\,e_2, \quad T = e_1 + \frac{b}{a}\,e_2\)$

Then:

$\(S \otimes T = a\,e_1\otimes e_1 + b\,e_1\otimes e_2 + c\,e_2\otimes e_1 + \frac{bc}{a}\,e_2\otimes e_2 = a\,e_1\otimes e_1 + b\,e_1\otimes e_2 + c\,e_2\otimes e_1 + d\,e_2\otimes e_2 = W\)$

  • If \(a = 0\): From \(ad - bc = 0\), we get \(bc = 0\).

  • \(b = 0\) and \(c = 0\): \(W = d\,e_2\otimes e_2 = (e_2) \otimes (d\,e_2)\), which is decomposable.

  • \(b = 0\) and \(c \neq 0\): \(bc = 0\) is automatically satisfied. Since \(a = 0, b = 0\), we have \(W = c\,e_2\otimes e_1 + d\,e_2\otimes e_2 = e_2 \otimes (c\,e_1 + d\,e_2)\), which is decomposable.
  • \(b \neq 0\) and \(c = 0\): Similarly, \(W = b\,e_1\otimes e_2 + d\,e_2\otimes e_2 = (b\,e_1 + d\,e_2) \otimes e_2\), which is decomposable.

In all cases, \(W\) is decomposable.

Conclusion

\[\boxed{W \text{ is decomposable} \iff ad - bc = 0}\]

Verification: For the example in the text, \(W = e_1\otimes e_1 + e_2\otimes e_2\), we have \(a = 1, b = 0, c = 0, d = 1\), so \(ad - bc = 1 \neq 0\). This is consistent with it being indecomposable. ✓

M-2. Component Representation of Bilinear Maps

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Problem: Calculations for a \(\binom{0}{2}\) tensor \(f\) with components \(f_{11}=2, f_{12}=1, f_{21}=1, f_{22}=3\).

(a) Express \(f(\vec{A}, \vec{B})\) in terms of components

Since \(f\) is a bilinear map, for \(\vec{A} = A^i e_i\) and \(\vec{B} = B^j e_j\):

\[f(\vec{A}, \vec{B}) = f(A^i e_i,\; B^j e_j)\]

By linearity in the first argument:

\[= A^i\,f(e_i,\; B^j e_j)\]

By linearity in the second argument:

\[= A^i B^j\,f(e_i, e_j)\]

Since \(f(e_i, e_j) = f_{ij}\):

\[\boxed{f(\vec{A}, \vec{B}) = A^i B^j f_{ij} = f_{ij}\,A^i B^j}\]

(Using Einstein summation convention, summing over \(i\) and \(j\).)

(b) Calculation of the specific value

From \(\vec{A} = e_1 - 2e_2\), we have \(A^1 = 1, A^2 = -2\). From \(\vec{B} = 3e_1 + e_2\), we have \(B^1 = 3, B^2 = 1\).

\[f(\vec{A}, \vec{B}) = A^1 B^1 f_{11} + A^1 B^2 f_{12} + A^2 B^1 f_{21} + A^2 B^2 f_{22}\]
\[= (1)(3)(2) + (1)(1)(1) + (-2)(3)(1) + (-2)(1)(3)\]
\[= 6 + 1 - 6 - 6 = \boxed{-5}\]

(c) Verification of symmetry

Compare \(f(\vec{A}, \vec{B}) = A^i B^j f_{ij}\) with \(f(\vec{B}, \vec{A}) = B^i A^j f_{ij}\).

In \(f(\vec{B}, \vec{A})\), relabeling the dummy indices \(i \to j\), \(j \to i\):

\[f(\vec{B}, \vec{A}) = B^i A^j f_{ij} = B^j A^i f_{ji} = A^i B^j f_{ji}\]

Meanwhile, \(f(\vec{A}, \vec{B}) = A^i B^j f_{ij}\).

For \(f(\vec{A}, \vec{B}) = f(\vec{B}, \vec{A})\) to hold for arbitrary \(\vec{A}, \vec{B}\), we need \(A^i B^j f_{ij} = A^i B^j f_{ji}\) to hold for arbitrary \(A^i, B^j\), which is equivalent to \(f_{ij} = f_{ji}\).

Checking explicitly:

\[f_{11} = 2 = f_{11}, \quad f_{12} = 1 = f_{21}, \quad f_{21} = 1 = f_{12}, \quad f_{22} = 3 = f_{22}\]

Since \(f_{ij} = f_{ji}\) holds for all \(i, j\), we conclude that \(f(\vec{A}, \vec{B}) = f(\vec{B}, \vec{A})\) holds for arbitrary \(\vec{A}, \vec{B}\). \(\square\)

Verification: Directly computing \(f(\vec{B}, \vec{A})\) from part (b). \(B^1 = 3, B^2 = 1, A^1 = 1, A^2 = -2\).

\[f(\vec{B}, \vec{A}) = (3)(1)(2) + (3)(-2)(1) + (1)(1)(1) + (1)(-2)(3) = 6 - 6 + 1 - 6 = -5\]

\(f(\vec{A}, \vec{B}) = f(\vec{B}, \vec{A}) = -5\). ✓

M-3. Rank Addition via Tensor Product

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Problem: Show using component notation that the tensor product of \(S \in T^r(V)\) and \(T \in T^m(V)\) is an element of \(T^{r+m}(V)\).

Solution

Express \(S\) and \(T\) in component form:

\[S = S^{i_1 i_2 \cdots i_r}\,e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_r}\]
\[T = T^{j_1 j_2 \cdots j_m}\,e_{j_1} \otimes e_{j_2} \otimes \cdots \otimes e_{j_m}\]

We compute the tensor product \(S \otimes T\). Using rule (B.1) (scalar extraction) and rules (B.2), (B.3) (distributive laws):

\[S \otimes T = \left(S^{i_1 \cdots i_r}\,e_{i_1} \otimes \cdots \otimes e_{i_r}\right) \otimes \left(T^{j_1 \cdots j_m}\,e_{j_1} \otimes \cdots \otimes e_{j_m}\right)\]

First, by the distributive law, we can take the tensor product term by term for each basis tensor:

\[= S^{i_1 \cdots i_r}\,T^{j_1 \cdots j_m}\,\left(e_{i_1} \otimes \cdots \otimes e_{i_r}\right) \otimes \left(e_{j_1} \otimes \cdots \otimes e_{j_m}\right)\]

Here we used rule (B.1) to factor out the scalars \(S^{i_1 \cdots i_r}\) and \(T^{j_1 \cdots j_m}\) from the tensor product.

By the associativity of the tensor product, we can remove the parentheses:

\[= S^{i_1 \cdots i_r}\,T^{j_1 \cdots j_m}\,e_{i_1} \otimes \cdots \otimes e_{i_r} \otimes e_{j_1} \otimes \cdots \otimes e_{j_m}\]

Defining new components as

\[U^{i_1 \cdots i_r j_1 \cdots j_m} := S^{i_1 \cdots i_r}\,T^{j_1 \cdots j_m}\]

we obtain:

\[S \otimes T = U^{i_1 \cdots i_r j_1 \cdots j_m}\,e_{i_1} \otimes \cdots \otimes e_{i_r} \otimes e_{j_1} \otimes \cdots \otimes e_{j_m}\]

The right-hand side is expressed in terms of a basis consisting of tensor products of \(r + m\) basis vectors, with components \(U^{i_1 \cdots i_r j_1 \cdots j_m}\) carrying \(r + m\) upper indices. This is precisely the standard representation of an element of \(T^{r+m}(V)\).

\[\boxed{S \otimes T \in T^{r+m}(V), \quad (S \otimes T)^{i_1 \cdots i_r j_1 \cdots j_m} = S^{i_1 \cdots i_r}\,T^{j_1 \cdots j_m}}\]

\(\square\)

Verification: Dimensional consistency. \(T^r(V)\) has dimension \(n^r\), \(T^m(V)\) has dimension \(n^m\). \(T^{r+m}(V)\) has dimension \(n^{r+m} = n^r \cdot n^m\). The number of index combinations for the components \(U^{i_1 \cdots i_r j_1 \cdots j_m}\) is \(n^r \cdot n^m = n^{r+m}\), which matches. ✓

M-4. \(T^0(V)\) and Scalars

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Problem: Show that elements of \(T^0(V)\) correspond to scalars, and that \(\lambda \otimes S = \lambda S\).

Solution

\(T^0(V)\) corresponds to scalars:

Setting \(r = 0\) in Definition B.1, the basis of \(T^0(V)\) is \(e_{i_1} \otimes \cdots \otimes e_{i_r}\) with \(r = 0\) tensor products, i.e., the "empty tensor product." This consists of only a single symbol (written as the unit element \(\mathbf{1}\)). The dimension of \(T^0(V)\) is \(n^0 = 1\).

Any element of \(T^0(V)\) can be written in the form \(\lambda \cdot \mathbf{1}\) (\(\lambda \in \mathbb{R}\)). Therefore \(T^0(V)\) is isomorphic to the field of real numbers \(\mathbb{R}\), and elements of \(T^0(V)\) correspond to scalars (real numbers).

Proof that \(\lambda \otimes S = \lambda S\):

Let \(\lambda \in T^0(V)\) (a scalar) and \(S \in T^r(V)\). Express \(S\) in component form as \(S = S^{i_1 \cdots i_r}\,e_{i_1} \otimes \cdots \otimes e_{i_r}\).

We use the tensor product rule (B.1): \(k(S \otimes T) = (kS) \otimes T = S \otimes (kT)\).

Since \(\lambda\) is a scalar, applying rule (B.1) to \(\lambda \otimes S\):

\[\lambda \otimes S = \lambda \otimes \left(S^{i_1 \cdots i_r}\,e_{i_1} \otimes \cdots \otimes e_{i_r}\right)\]

By the distributive law and (B.1):

\[= S^{i_1 \cdots i_r}\,\lambda \otimes \left(e_{i_1} \otimes \cdots \otimes e_{i_r}\right)\]

Here, \(\lambda \otimes (e_{i_1} \otimes \cdots \otimes e_{i_r})\) allows the scalar \(\lambda\) to be factored out in front of the basis tensor, in the sense of rule (B.1): \(k(S \otimes T) = (kS) \otimes T\). That is, the tensor product of an element \(\lambda\) of \(T^0(V)\) with a basis element of \(T^r(V)\) is an element of \(T^{0+r}(V) = T^r(V)\):

\[\lambda \otimes (e_{i_1} \otimes \cdots \otimes e_{i_r}) = \lambda\,(e_{i_1} \otimes \cdots \otimes e_{i_r})\]

Therefore:

\[\lambda \otimes S = S^{i_1 \cdots i_r}\,\lambda\,(e_{i_1} \otimes \cdots \otimes e_{i_r}) = \lambda\,S^{i_1 \cdots i_r}\,(e_{i_1} \otimes \cdots \otimes e_{i_r}) = \lambda S\]
\[\boxed{\lambda \otimes S = \lambda S \in T^r(V)}\]

\(\square\)

Verification: Rank consistency: the tensor product of an element of \(T^0(V)\) and an element of \(T^r(V)\) is an element of \(T^{0+r}(V) = T^r(V)\). The scalar multiple \(\lambda S\) is indeed an element of \(T^r(V)\). ✓

M-5. Simple \(S \otimes T\) Expansion

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Solution Strategy

Expand the tensor product of \(S = e_1 + 2e_2\) and \(T = 3e_1 - e_2\) using the distributive law, and express the result in terms of the basis \(\{e_i \otimes e_j\}\) of \(T^2(V)\).

Calculation

\[ S \otimes T = (e_1 + 2e_2) \otimes (3e_1 - e_2) \]

Applying the distributive law:

\[ = e_1 \otimes (3e_1 - e_2) + 2e_2 \otimes (3e_1 - e_2) \]
\[ = 3\,e_1 \otimes e_1 - e_1 \otimes e_2 + 6\,e_2 \otimes e_1 - 2\,e_2 \otimes e_2 \]
\[ \boxed{S \otimes T = 3\,e_1 \otimes e_1 - e_1 \otimes e_2 + 6\,e_2 \otimes e_1 - 2\,e_2 \otimes e_2} \]

Written in component form, \((S \otimes T)^{ij}\) is

\[ (S \otimes T)^{ij} = S^i T^j = \begin{pmatrix} 3 & -1 \\ 6 & -2 \end{pmatrix} \]

Verification

We check the decomposability condition \(ad - bc = 0\): with \(a = 3\), \(b = -1\), \(c = 6\), \(d = -2\), we have \(ad - bc = 3 \times (-2) - (-1) \times 6 = -6 + 6 = 0\). ✓ This confirms that the tensor is decomposable. \(\square\)

M-6. Dimension of Tensor Spaces in 3-Dimensional Space

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Solution Strategy

Use the fact that when \(V\) is \(n\)-dimensional, the dimension of \(T^r(V)\) is \(n^r\), and find the dimensions of \(T^2(V)\) and \(T^3(V)\) for the case \(n = 3\).

Calculation

When \(V\) is an \(n\)-dimensional linear space, the basis of \(T^r(V)\) is

\[ \{e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_r}\}_{i_1, i_2, \ldots, i_r = 1, \ldots, n} \]

Since each index takes \(n\) values, the number of basis elements (= dimension) is \(n^r\).

When \(V\) is 3-dimensional (\(n = 3\)):

Dimension of \(T^2(V)\):

\[ \dim T^2(V) = 3^2 = 9 \]

The basis consists of 9 elements \(e_i \otimes e_j\) (\(i, j = 1, 2, 3\)): \(e_1 \otimes e_1\), \(e_1 \otimes e_2\), \(e_1 \otimes e_3\), \(e_2 \otimes e_1\), \(e_2 \otimes e_2\), \(e_2 \otimes e_3\), \(e_3 \otimes e_1\), \(e_3 \otimes e_2\), \(e_3 \otimes e_3\).

Dimension of \(T^3(V)\):

\[ \dim T^3(V) = 3^3 = 27 \]

The basis consists of 27 elements \(e_i \otimes e_j \otimes e_k\) (\(i, j, k = 1, 2, 3\)).

\[ \boxed{\dim T^2(V) = 9, \quad \dim T^3(V) = 27} \]

Verification

In general, \(\dim T^r(V) = n^r\) grows exponentially. For \(n = 3\), the dimensions increase as \(T^0 = 1\), \(T^1 = 3\), \(T^2 = 9\), \(T^3 = 27\), \(T^4 = 81\). In 4-dimensional spacetime (\(n = 4\)), the Riemann tensor is an element of \(T^4\) and has \(4^4 = 256\) components, but due to symmetries the number of independent components reduces to 20. \(\square\)

M-7. Non-decomposability of Antisymmetric Tensors

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Solution Strategy

We show that \(W = e_1 \otimes e_2 - e_2 \otimes e_1\) cannot be written as a decomposable tensor (in the form \(S \otimes T\)) by using the criterion \(ad - bc = 0\) from Problem B.9.

Calculation

The components of \(W = e_1 \otimes e_2 - e_2 \otimes e_1\) are

\[ W^{11} = 0, \quad W^{12} = 1, \quad W^{21} = -1, \quad W^{22} = 0 \]

That is, \(a = 0\), \(b = 1\), \(c = -1\), \(d = 0\).

From the result of Problem B.9, the necessary and sufficient condition for \(W\) to be a decomposable tensor (\(W = S \otimes T\)) is

\[ ad - bc = 0 \]

Computing explicitly,

\[ ad - bc = 0 \times 0 - 1 \times (-1) = 0 + 1 = 1 \neq 0 \]
\[ \boxed{ad - bc = 1 \neq 0 \quad \Longrightarrow \quad e_1 \otimes e_2 - e_2 \otimes e_1 \text{ is not a decomposable tensor}} \]

Physical significance: \(e_1 \otimes e_2 - e_2 \otimes e_1\) is a typical example of an antisymmetric tensor and corresponds to the "exterior product" of two vectors. In general, nontrivial antisymmetric tensors cannot be expressed as decomposable tensors. This is related to the fact that the electromagnetic field tensor \(F_{\mu\nu}\) (antisymmetric) generally cannot be written in the form of a single \(A_\mu \otimes B_\nu\).

Verification

If we could write \(W = S \otimes T = (\alpha e_1 + \beta e_2) \otimes (\gamma e_1 + \delta e_2)\), then \(W^{11} = \alpha\gamma = 0\), \(W^{12} = \alpha\delta = 1\), \(W^{21} = \beta\gamma = -1\), \(W^{22} = \beta\delta = 0\). From \(\alpha\gamma = 0\), either \(\alpha = 0\) or \(\gamma = 0\). If \(\alpha = 0\), then \(\alpha\delta = 0 \neq 1\), a contradiction. If \(\gamma = 0\), then \(\beta\gamma = 0 \neq -1\), a contradiction. Therefore, the indecomposability is also confirmed directly. \(\square\)

M-8. Evaluation of the Identity Bilinear Form

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Solution Strategy

Read off the components \(f_{ij} = f(e_i, e_j)\) of the \(\binom{0}{2}\) tensor \(f\), and compute using multilinearity \(f(\vec{A}, \vec{B}) = A^i B^j f_{ij}\).

Calculation

From the given conditions, the components of \(f\) are

\[ f_{11} = f(e_1, e_1) = 1, \quad f_{12} = f(e_1, e_2) = 0, \quad f_{21} = f(e_2, e_1) = 0, \quad f_{22} = f(e_2, e_2) = 1 \]

That is, \(f_{ij} = \delta_{ij}\) (Kronecker delta, the identity matrix).

For \(\vec{A} = 3e_1 + 2e_2\) (\(A^1 = 3\), \(A^2 = 2\)) and \(\vec{B} = -e_1 + 4e_2\) (\(B^1 = -1\), \(B^2 = 4\)):

\[ f(\vec{A}, \vec{B}) = A^i B^j f_{ij} = \sum_{i=1}^{2}\sum_{j=1}^{2} A^i B^j f_{ij} \]

Since \(f_{ij} = \delta_{ij}\), all terms with \(i \neq j\) vanish:

\[ f(\vec{A}, \vec{B}) = A^1 B^1 f_{11} + A^2 B^2 f_{22} = A^1 B^1 + A^2 B^2 \]
\[ = 3 \times (-1) + 2 \times 4 = -3 + 8 = 5 \]
\[ \boxed{f(\vec{A}, \vec{B}) = 5} \]

Verification

\(f_{ij} = \delta_{ij}\) corresponds to the Euclidean inner product (dot product). Therefore \(f(\vec{A}, \vec{B}) = \vec{A} \cdot \vec{B} = 3 \times (-1) + 2 \times 4 = 5\). ✓

Also, \(f(\vec{B}, \vec{A}) = B^1 A^1 + B^2 A^2 = (-1) \times 3 + 4 \times 2 = 5 = f(\vec{A}, \vec{B})\), which is consistent with \(f\) being a symmetric tensor (\(f_{ij} = f_{ji}\)). \(\square\)

Advanced

A-1. Symmetric and Antisymmetric Decomposition

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Problem: Symmetric-antisymmetric decomposition of a rank-2 tensor.

(a) Unique decomposition \(W = W^{(S)} + W^{(A)}\)

Proof of existence:

For any \(W \in T^2(V)\), write \(W = W^{ij}\,e_i \otimes e_j\). By definition:

\[W^{(S)} = \frac{1}{2}(W^{ij} + W^{ji})\,e_i \otimes e_j, \qquad W^{(A)} = \frac{1}{2}(W^{ij} - W^{ji})\,e_i \otimes e_j\]

Computing the sum:

\[W^{(S)} + W^{(A)} = \frac{1}{2}(W^{ij} + W^{ji})\,e_i \otimes e_j + \frac{1}{2}(W^{ij} - W^{ji})\,e_i \otimes e_j\]
\[= \frac{1}{2}\left[(W^{ij} + W^{ji}) + (W^{ij} - W^{ji})\right]\,e_i \otimes e_j = \frac{1}{2} \cdot 2W^{ij}\,e_i \otimes e_j = W^{ij}\,e_i \otimes e_j = W\]

Therefore \(W = W^{(S)} + W^{(A)}\) holds.

Verification that \(W^{(S)}\) is symmetric: The \((ij)\) component of \(W^{(S)}\) is \(\frac{1}{2}(W^{ij} + W^{ji})\). The \((ji)\) component is \(\frac{1}{2}(W^{ji} + W^{ij})\). These are equal, so it is symmetric.

Verification that \(W^{(A)}\) is antisymmetric: The \((ij)\) component of \(W^{(A)}\) is \(\frac{1}{2}(W^{ij} - W^{ji})\). The \((ji)\) component is \(\frac{1}{2}(W^{ji} - W^{ij}) = -\frac{1}{2}(W^{ij} - W^{ji})\). Therefore it is antisymmetric.

Proof of uniqueness:

Suppose \(W = P + Q\) is a decomposition (\(P\) symmetric, \(Q\) antisymmetric). In components, \(W^{ij} = P^{ij} + Q^{ij}\).

Since \(P^{ij} = P^{ji}\) and \(Q^{ij} = -Q^{ji}\), swapping indices \(i, j\) gives:

\[W^{ji} = P^{ji} + Q^{ji} = P^{ij} - Q^{ij}\]

Solving the system \(W^{ij} = P^{ij} + Q^{ij}\) and \(W^{ji} = P^{ij} - Q^{ij}\) simultaneously:

\[P^{ij} = \frac{1}{2}(W^{ij} + W^{ji}), \qquad Q^{ij} = \frac{1}{2}(W^{ij} - W^{ji})\]

This is precisely the definition of \(W^{(S)}\) and \(W^{(A)}\), so the decomposition is unique. \(\square\)

(b) Subspace property and dimensions

\(\mathrm{Sym}^2(V)\) is a subspace:

  1. Zero element: \(0^{ij} = 0 = 0^{ji}\), so \(0 \in \mathrm{Sym}^2(V)\).
  2. Closure under addition: If \(P, Q \in \mathrm{Sym}^2(V)\), then \((P+Q)^{ij} = P^{ij} + Q^{ij} = P^{ji} + Q^{ji} = (P+Q)^{ji}\).
  3. Closure under scalar multiplication: If \(P \in \mathrm{Sym}^2(V)\), \(\lambda \in \mathbb{R}\), then \((\lambda P)^{ij} = \lambda P^{ij} = \lambda P^{ji} = (\lambda P)^{ji}\).

Therefore \(\mathrm{Sym}^2(V)\) is a subspace of \(T^2(V)\).

\(\mathrm{Alt}^2(V)\) is a subspace:

  1. Zero element: \(0^{ij} = 0 = -0 = -0^{ji}\), so \(0 \in \mathrm{Alt}^2(V)\).
  2. Closure under addition: If \(P, Q \in \mathrm{Alt}^2(V)\), then \((P+Q)^{ij} = P^{ij} + Q^{ij} = -P^{ji} + (-Q^{ji}) = -(P+Q)^{ji}\).
  3. Closure under scalar multiplication: If \(P \in \mathrm{Alt}^2(V)\), \(\lambda \in \mathbb{R}\), then \((\lambda P)^{ij} = \lambda P^{ij} = \lambda(-P^{ji}) = -(\lambda P)^{ji}\).

Therefore \(\mathrm{Alt}^2(V)\) is also a subspace of \(T^2(V)\). \(\square\)

Dimension calculation:

When \(V\) is \(n\)-dimensional, the number of independent components of a symmetric tensor \(S^{ij} = S^{ji}\) equals the number of pairs \((i,j)\) with \(i \leq j\). This is the number of combinations choosing 2 from \(n\) with repetition:

\[\dim \mathrm{Sym}^2(V) = \binom{n+1}{2} = \frac{n(n+1)}{2}\]

For an antisymmetric tensor \(A^{ij} = -A^{ji}\), the diagonal components vanish \(A^{ii} = 0\) (from \(A^{ii} = -A^{ii}\)). The number of independent components equals the number of pairs with \(i < j\):

\[\dim \mathrm{Alt}^2(V) = \binom{n}{2} = \frac{n(n-1)}{2}\]

Verification: \(\dim \mathrm{Sym}^2(V) + \dim \mathrm{Alt}^2(V) = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2 + n + n^2 - n}{2} = n^2 = \dim T^2(V)\). ✓

This is consistent with the direct sum decomposition \(T^2(V) = \mathrm{Sym}^2(V) \oplus \mathrm{Alt}^2(V)\) from part (a).

(c) Number of independent components in 4-dimensional spacetime

Substituting \(n = 4\).

Metric tensor \(g_{\mu\nu}\) (symmetric):

\[\frac{n(n+1)}{2} = \frac{4 \times 5}{2} = \boxed{10 \text{ components}}\]

Electromagnetic field tensor \(F_{\mu\nu}\) (antisymmetric):

\[\frac{n(n-1)}{2} = \frac{4 \times 3}{2} = \boxed{6 \text{ components}}\]

Verification of the sum:

\[10 + 6 = 16 = 4^2 = n^2 \quad \checkmark\]

This matches the dimension \(n^2 = 16\) of \(T^2(V)\), reflecting the fact that any rank-2 tensor (16 independent components) is uniquely decomposed into a symmetric part (10 components) and an antisymmetric part (6 components).

Physical remarks: The 6 independent components of \(F_{\mu\nu}\) correspond to the 3 components of the electric field \((E_x, E_y, E_z)\) and the 3 components of the magnetic field \((B_x, B_y, B_z)\). The 10 independent components of \(g_{\mu\nu}\) correspond to the fact that the Einstein equations \(G_{\mu\nu} = 8\pi T_{\mu\nu}\) form a system of 10 independent equations.

A-2. Tensor Representation via Dual Spaces

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Problem: The relationship between tensor products of dual spaces and \(\binom{0}{N}\) tensors.

(a) Basis and dimension of \(V^* \otimes V^*\)

Since the basis of \(V^*\) is \(\{e^1, e^2, \ldots, e^n\}\) (\(n\) elements), the basis of \(V^* \otimes V^*\) is:

\[\{e^i \otimes e^j \mid i, j = 1, 2, \ldots, n\}\]

The number of basis elements is \(n \times n = n^2\).

\[\boxed{\text{Basis: } \{e^i \otimes e^j\}_{i,j=1}^n, \qquad \dim(V^* \otimes V^*) = n^2}\]

(b) Verification of \(f = f_{ij}\,e^i \otimes e^j\)

Assuming \(f = f_{ij}\,e^i \otimes e^j\), we verify that it returns the correct value for arbitrary vectors \(\vec{A} = A^k e_k\) and \(\vec{B} = B^l e_l\).

Using the definition of the tensor product action \((e^i \otimes e^j)(\vec{A}, \vec{B}) := e^i(\vec{A})\,e^j(\vec{B})\):

\[f(\vec{A}, \vec{B}) = \left(f_{ij}\,e^i \otimes e^j\right)(\vec{A}, \vec{B})\]

By linearity:

\[= f_{ij}\,(e^i \otimes e^j)(\vec{A}, \vec{B}) = f_{ij}\,e^i(\vec{A})\,e^j(\vec{B})\]

From the definition of the dual basis \(e^i(e_k) = \delta^i{}_k\) and the linearity of \(\vec{A} = A^k e_k\):

\[e^i(\vec{A}) = e^i(A^k e_k) = A^k\,e^i(e_k) = A^k\,\delta^i{}_k = A^i\]

Similarly:

\[e^j(\vec{B}) = B^j\]

Therefore:

\[f(\vec{A}, \vec{B}) = f_{ij}\,A^i\,B^j = A^i B^j f_{ij}\]

This is in complete agreement with the component representation of a \(\binom{0}{2}\) tensor derived in S2(a): \(f(\vec{A}, \vec{B}) = f_{ij} A^i B^j\).

Conversely, for any \(\binom{0}{2}\) tensor \(f\) (a bilinear map \(f: V \times V \to \mathbb{R}\)), if we define \(f_{ij} := f(e_i, e_j)\), then from the calculation above, \(f\) can be identified with the element \(f_{ij}\,e^i \otimes e^j\) of \(V^* \otimes V^*\).

\[\boxed{f = f_{ij}\,e^i \otimes e^j \in V^* \otimes V^*, \quad f(\vec{A}, \vec{B}) = f_{ij}\,A^i B^j \text{ is reproduced}}\]

\(\square\)

(c) Generalization and connection between the two approaches

The space of \(\binom{0}{2}\) tensors \(\cong V^* \otimes V^*\):

From the result of (b), the totality of \(\binom{0}{2}\) tensors (bilinear maps on \(V \times V\)) is isomorphic to \(V^* \otimes V^*\). The map is given by:

\[f \mapsto f_{ij}\,e^i \otimes e^j, \qquad f_{ij} = f(e_i, e_j)\]

and this is a linear isomorphism (independent of the choice of basis).

Generalization to \(\binom{0}{N}\) tensors:

Similarly, for a \(\binom{0}{N}\) tensor (a multilinear map taking \(N\) vectors as arguments, \(f: \underbrace{V \times \cdots \times V}_{N} \to \mathbb{R}\)):

\[f = f_{i_1 i_2 \cdots i_N}\,e^{i_1} \otimes e^{i_2} \otimes \cdots \otimes e^{i_N}\]

Here \(f_{i_1 \cdots i_N} = f(e_{i_1}, \ldots, e_{i_N})\), and the action is:

\[(e^{i_1} \otimes \cdots \otimes e^{i_N})(\vec{A}_1, \ldots, \vec{A}_N) := e^{i_1}(\vec{A}_1) \cdots e^{i_N}(\vec{A}_N) = A_1^{i_1} \cdots A_N^{i_N}\]

Therefore:

\[f(\vec{A}_1, \ldots, \vec{A}_N) = f_{i_1 \cdots i_N}\,A_1^{i_1} \cdots A_N^{i_N}\]

This gives:

\[\boxed{\text{The space of } \binom{0}{N}\text{ tensors} \cong \underbrace{V^* \otimes \cdots \otimes V^*}_{N} =: T_N(V)}\]

Connection between the two approaches (generalization of Section B.11 in the main text):

There are two equivalent definitions of tensors:

Approach Contravariant tensors Covariant tensors
Algebraic (tensor product space) \(T^r(V) = \underbrace{V \otimes \cdots \otimes V}_{r}\) \(T_N(V) = \underbrace{V^* \otimes \cdots \otimes V^*}_{N}\)
Functional (multilinear maps) \(\underbrace{V^* \times \cdots \times V^*}_{r} \to \mathbb{R}\) \(\underbrace{V \times \cdots \times V}_{N} \to \mathbb{R}\)
  • Contravariant tensors \(T^r(V)\): Elements of the space formed by taking the tensor product of \(V\) with itself \(r\) times. Components carry \(r\) upper indices. Equivalently, multilinear maps taking \(r\) one-forms as arguments.
  • Covariant tensors \(T_N(V)\): Elements of the space formed by taking the tensor product of \(V^*\) with itself \(N\) times. Components carry \(N\) lower indices. Equivalently, multilinear maps taking \(N\) vectors as arguments.

Furthermore, general mixed tensors (of type \(\binom{r}{N}\)) are elements of:

\[T^r{}_N(V) = \underbrace{V \otimes \cdots \otimes V}_{r} \otimes \underbrace{V^* \otimes \cdots \otimes V^*}_{N}\]

with components \(T^{i_1 \cdots i_r}{}_{j_1 \cdots j_N}\) carrying \(r\) upper and \(N\) lower indices. Equivalently, they are multilinear maps taking \(r\) one-forms and \(N\) vectors as arguments.

This correspondence is the general expression of the fact that the two approaches described in Section B.11 of the main text—"elements of a tensor product space" and "multilinear maps"—describe the same mathematical object. \(\square\)

Verification: Dimensional consistency. The dimension of \(T_N(V) = (V^*)^{\otimes N}\) is \((\dim V^*)^N = n^N\). On the other hand, the number of components \(f_{i_1 \cdots i_N}\) of a multilinear map taking \(N\) vectors as arguments is \(n^N\). They agree. ✓