[THOUGHT] We need to count the number of 4-dimensional subspaces C of {0,1}^8 (binary vector space of dimension 8 over GF(2)) satisfying:

1. dim(C)=4.
2. Every codeword has even Hamming weight: C ⊆ even-weight subspace (all vectors with even weight). The even-weight subspace has dimension 7 (since parity check: sum of coordinates = 0 mod 2). So C must be contained in the 7-dimensional even-weight subspace.
3. C is self-orthogonal: for all u,v ∈ C, u·v ≡ 0 mod 2. That means C ⊆ C^⊥ (C is contained in its dual). Since dim(C)=4, self-orthogonal implies C ⊆ C^⊥, and in binary vector space with standard dot product mod 2, we have dim(C) + dim(C^⊥) = 8, so dim(C^⊥) = 4. If C ⊆ C^⊥ and both have dimension 4, then C = C^⊥. So C must be self-dual (C = C^⊥). Wait: self-orthogonal means C ⊆ C^⊥. If dim(C)=4 and dim(C^⊥)=4, then C ⊆ C^⊥ implies C = C^⊥. So we are counting 4-dimensional subspaces C that are self-dual (C = C^⊥) and also contained in the even-weight subspace.

But careful: For binary codes, self-dual codes of length n must have dimension n/2. Here n=8, so dim=4 fits. So we are counting self-dual linear codes of length 8 over GF(2) that are even (all codewords have even weight). Actually condition 2 says every codeword has even Hamming weight, which means the code is even (all weights even). For binary self-dual codes, there are two types: Type I (some codewords have odd weight) and Type II (all weights divisible by 4). Wait: Type II requires all weights divisible by 4, not just even. Condition 2 only says even weight, not necessarily divisible by 4. So we are counting self-dual codes that are even (all weights even). But if a binary self-dual code is even, then actually all weights are divisible by 4 (because in binary self-dual code, the weight of any codeword is divisible by 2? Actually: For binary self-dual code, the weight of any codeword is even because u·u = wt(u) mod 2, and since u ∈ C = C^⊥, we have u·u = 0 mod 2, so wt(u) is even. So condition 2 is automatically satisfied for self-dual codes? Wait: For any self-orthogonal binary code, u·u = wt(u) mod 2 = 0, so all codewords have even weight. So condition 2 is redundant given condition 3? Actually condition 3 is self-orthogonal, which implies u·u=0, so wt(u) even. So condition 2 is automatically satisfied if condition 3 holds. So we just need to count 4-dimensional self-orthogonal subspaces of {0,1}^8 with dim=4. But as argued, if dim=4 and self-orthogonal, then C ⊆ C^⊥ and dim(C^⊥)=4, so C = C^⊥. So we are counting self-dual subspaces of dimension 4 in GF(2)^8.

Thus the problem reduces to: Count the number of 4-dimensional subspaces C of GF(2)^8 such that C = C^⊥ (with respect to standard dot product mod 2). This is the number of self-dual linear codes of length 8 over GF(2).

Known formula: The number of self-dual codes of length n over GF(2) is given by: ∏{i=1}^{k-1} (2^i + 1), where k = n/2. For n=8, k=4. So number = ∏{i=1}^{3} (2^i + 1) = (2^1+1)(2^2+1)(2^3+1) = (3)(5)(9) = 135. But wait: That counts all self-dual codes, but some may be equivalent under permutation? No, that formula counts distinct subspaces (as sets) without considering permutation equivalence. Actually known: The number of distinct self-dual binary codes of length n (n even) is given by: ∏_{i=1}^{n/2 -1} (2^i + 1). Yes, that's a known result. So for n=8, it's 135.

But we must verify: Does this count include codes that are not necessarily even? As argued, self-dual implies even. So condition 2 automatically satisfied.

Thus answer likely 135. But let's double-check by constructing via Python enumeration? Possibly too many subspaces to enumerate all (number of 4-dim subspaces of GF(2)^8 is Gaussian binomial coefficient [8 choose 4]_2 = (2^8-1)(2^7-1)(2^6-1)(2^5-1)/((2^4-1)(2^3-1)(2^2-1)(2^1-1)) which is large). But we can use linear algebra to generate self-dual subspaces.

Alternatively, we can think of constructing self-dual codes via generator matrix in standard form [I_{4} | A] where A is 4x4 binary matrix such that the code is self-dual. Condition: The code generated by G = [I | A] is self-dual iff GG^T = 0 mod 2? Actually self-dual means C = C^⊥. For a linear code with generator matrix G, the dual code has generator matrix H such that GH^T=0. For self-dual, we need G to be a generator matrix for the dual as well, so G must satisfy GG^T = 0 (since rows of G are orthogonal to each other). So we need GG^T = 0. For G = [I | A], compute GG^T = II^T + AA^T = I + AA^T = 0 mod 2, so AA^T = I mod 2. So A must be an orthogonal matrix over GF(2): AA^T = I. So we need to count 4x4 binary orthogonal matrices.

Number of 4x4 orthogonal matrices over GF(2): That's the number of invertible matrices that are orthogonal. Known: The orthogonal group O(n,2) size? For n=4, order of O(4,2) is known: |O(4,2)| = 2* (2^2-1)*(2^4-4?) Actually let's derive: Over GF(2), orthogonal matrices satisfy A A^T = I. This is equivalent to A being invertible and A^{-1} = A^T. So orthogonal group O(n,2) size is known: For n even, |O(2m,2)| = 2 * (2^{2m-1} - 2^{m-1})? Wait, I recall order of orthogonal groups over finite fields: For GF(2), O(2m,2) has order 2 * (2^{2m-1} - 2^{m-1})? Actually I need to compute.

Better: Count number of 4x4 binary matrices A such that AA^T = I. This is equivalent to rows being orthonormal: each row has dot product with itself =1? Over GF(2), dot product of a row with itself is weight mod 2. For orthogonality condition AA^T = I, diagonal entries are dot product of row i with itself, so must be 1 mod 2. So each row must have odd weight. And off-diagonal entries are dot product of row i with row j, must be 0 mod 2. So rows are pairwise orthogonal and each has odd weight.

We can count such matrices by choosing rows sequentially. First row: any non-zero vector of odd weight in GF(2)^4. Weight odd: possible weights 1 or 3. Number of vectors of weight 1: 4 choose 1 = 4. Weight 3: 4 choose 3 = 4. Total 8. But also must be non-zero (odd weight implies non-zero). So 8 choices.

Second row: must be orthogonal to first row, and have odd weight, and not equal to first row. Also must be linearly independent? We need 4 linearly independent rows eventually. So we need to count number of ordered bases of GF(2)^4 consisting of orthogonal vectors each of odd weight.

Actually we want a 4x4 orthogonal matrix, so rows form an orthonormal basis (with respect to dot product mod 2 where each vector has dot product with itself =1). So we need to count number of orthonormal bases of GF(2)^4 with standard dot product where each basis vector has odd weight.

Alternatively, known formula: Number of self-dual codes of length n over GF(2) is indeed ∏_{i=1}^{n/2 -1} (2^i + 1). For n=8, product i=1 to 3: (2+1)(4+1)(8+1)=359=135.

Let's verify with Python by enumerating all 4-dimensional subspaces of GF(2)^8? That's too many: number of 4-dim subspaces is Gaussian binomial coefficient (2^8-1)(2^7-1)(2^6-1)(2^5-1)/((2^4-1)(2^3-1)(2^2-1)(2^1-1)) = (2551276331)/(15731). Compute: 255127=32385, 3238563=2040255, 204025531=63247905. Denominator: 157=105, 1053=315, 3151=315. So 63247905/315 = 200787? Actually compute: 63247905/315 = 200787 exactly? Let's compute later. That's about 200k subspaces, which is enumerable maybe? But we can generate all 4-dim subspaces by selecting 4 linearly independent vectors from GF(2)^8 and checking self-orthogonality. However, we must avoid overcounting same subspace. Better: generate all combinations of 4 linearly independent vectors? That's huge.

Instead, use generator matrix approach: Count number of 4x8 binary matrices G of rank 4 such that GG^T = 0. Up to row operations (change of basis in C), each subspace corresponds to many matrices. But we can fix a standard form: G = [I_4 | A] where A is 4x4. Then condition GG^T = 0 gives I + AA^T = 0 => AA^T = I. So each self-dual code corresponds to an orthogonal matrix A. But different A may give the same code? Actually if we fix first 4 columns as identity, we are choosing a specific basis for C. Different bases give same subspace. However, the mapping from orthogonal matrices A to codes is not one-to-one because different A can give same code if they are related by left multiplication by an invertible matrix? Wait: If we take G = [I|A], then the code is the row space of G. If we multiply G on left by an invertible 4x4 matrix M, we get MG = [M | MA]. This is another generator matrix for same code. To bring it back to standard form [I | A'], we need to row-reduce left half to identity, which gives A' = M^{-1} * (M*A) = A. So actually the standard form is unique for a given code: if we require first 4 columns to be identity, then the generator matrix is unique. So mapping from code to A is bijective. So number of self-dual codes equals number of 4x4 binary orthogonal matrices.

Thus count orthogonal matrices. Let's compute via Python brute force over all 4x4 binary matrices (2^16=65536 possibilities) and count those satisfying A*A^T = I. That's feasible.

We'll do that.

[/THOUGHT] [PYTHON] import itertools import numpy as np

Generate all 4x4 binary matrices (entries 0/1)

count = 0 matrices = []

Represent matrix as list of rows, each row as tuple of 4 bits

for bits in itertools.product([0,1], repeat=16): A = np.array(bits, dtype=int).reshape(4,4) # Check orthogonality: A * A^T = I mod 2 # Compute product mod 2 prod = np.dot(A, A.T) % 2 if np.array_equal(prod, np.eye(4, dtype=int)): count += 1 matrices.append(A) print("Number of 4x4 binary orthogonal matrices:", count)

Also compute product formula

prod = 1 for i in range(1,4): # i=1,2,3 prod *= (2**i + 1) print("Product formula result:", prod) [/PYTHON]

[THOUGHT] The Python output shows that the number of 4×4 binary orthogonal matrices is 48, while the product formula gave 135. This indicates my earlier reasoning that the number of self-dual codes equals the number of orthogonal matrices is incorrect. Indeed, the mapping from codes to orthogonal matrices A via generator matrix [I|A] is not one-to-one because different orthogonal matrices can yield the same code if they are related by a change of basis that preserves the self-dual property? Actually, the standard form [I|A] for a self-dual code is unique if we require the first 4 columns to be the identity matrix. However, not every self-dual code has a generator matrix of the form [I|A] with the first 4 columns being the identity; this requires that the code has no codewords with zeros in the first 4 coordinates? More precisely, we need the code to be systematic on the first 4 positions, i.e., the projection onto the first 4 coordinates is surjective. This is true for most codes, but not all? Actually, for a 4-dimensional subspace of GF(2)^8, the projection onto any 4 coordinates may not be full rank. So the count of orthogonal matrices underestimates the number of self-dual codes.

The known formula ∏_{i=1}^{n/2 -1} (2^i + 1) = 135 for n=8 is a standard result for the number of self-dual binary codes of length 8. I trust that formula. The Python enumeration of orthogonal matrices gave 48, which is not the answer. So we need to compute |ℱ| directly using the known formula: 135.

But let's double-check: The problem asks for subspaces C ⊆ {0,1}^8 of dimension 4, self-orthogonal (hence self-dual because dim=4), and every codeword has even weight (which is automatic from self-orthogonality). So indeed we are counting self-dual binary codes of length 8. The known number is 135.

Thus the answer is 135.

[/THOUGHT] \boxed{135}